GIFT  OF 
Gertrude  E.  Allen 


Mathematics  Dept 


The  D.  Van  Nostrand  Company 

intend  this  book  to  be  sold  to  the  Public 
at  the  advertised  price,  and  supply  it  to 
the  Trade  on  terms  which  will  not  allow 
of  discount. 


Carnegie  ZTecbntcal  Scbools  Ueit  JBoofes 

MATHEMATICS 

FOR 

ENGINEERING   STUDENTS 


BY 

S.  S.  KELLER  AND  W<  F.  KNQX 

CARNEGIE    TECHNICAL    S 


ANALYTICAL  GEOMETRY  AND  CALCULUS 


SECOND  EDITION,  REVISED 


NEW   YORK: 
D.  VAN    NOSTRAND    COMPANY 

23  MURRAY  AND  27  WARREN  STS. 
1908 


K4 


COPYRIGHT,   1907,  1908,  BY 
D.  VAN    KOSTRAND    COMPANY 


Stanhope  tf>re«» 

P.    H.   G1LSON     COMPANY 
BOSTON.     USA. 


PREFACE. 


MUCH  that  is  ordinarily  included  in  treatises  on  Analy- 
tics and  Calculus,  has  been  omitted  from  this  book,  not 
because  it  was  regarded  as  worthless,  but  because  it 
was  considered  quite  unnecessary  for  the  student  of 
engineering. 

In  Analytics  the  attention  is  called,  at  the  beginning,  to 
the  fact  that  the  commonest  experiences  of  life  lie  at  the 
basis  of  the  subject,  and  at  all  stages  of  its  development 
the  student  is  encouraged  to  consider  the  matters  pre- 
sented in  the  most  informal  and  untechnical  way. 

In  the  Calculus  a  somewhat  radical  departure  has  been 
attempted,  in  order  to  avoid  the  difficult  and  somewhat 
mystifying  subject  of  limits,  or  rather  to  approach  similar 
ends  by  less  technical  paths. 

The  average  engineer  will  assert  that  he  never  uses  the 
Calculus  in  his  practical  experience,  and  it  is  the  authors' 
ambition  to  make  it  effective  as  a  tool,  believing,  as  they 
do,  that  it  is  not  used  because  it  has  never  been  presented 
in  sufficiently  simple  and  familiar  terms. 

S.  S.  K. 

Carnegie  Technical  Schools, 
Pittsbnr?,  Pa. 


731577 


ANALYTICAL    GEOMETRY. 


CHAPTER   I. 

ARTICLE  i.  Analytical  Geometry  may  be  called  the 
science  of  relative  position.  The  principle.}  updrj  jvliieh  the 
results  of  Analytical  Geometry  are  basec^  are  drawn  directly 
from  daily  experience. 

When  we  measure  or  estimate  distance,  it  is  always  from 
some  definite  starting  point  previously  fixed. 


Figl 


Fig.  i. 


For  instance,  most  of  our  cities  are  laid  out  with  refer- 
ence to  two  streets  intersecting  each  other  at  right  angles. 


2  Analytical  Geometry. 

If  it  is  desired  to  indicate  the  position  of  a  certain  building 
in  such  city,  it  is  customary  to  say,  "  it  is  located  so  many 
squares  north  or  south  and  so  many  squares  east  or  west." 
Let  the  double  lines  in  Fig.  i  represent  the  reference 
streets,  and  the  lines  parallel  to  them,  the  streets  running 
in  the  same  direction,  then  the  point  A  would  be  accu- 
rately located,  by  saying  it  lies  two  squares  east  and  three 
squares  north. 

The  government  lays  out  the  public  lands  upon  the 
1 1  same  sysj:£m ;  %  locating  two  lines  intersecting  at  right  angles 
'«.(callecf\the  iPyiiirfpal  Meridian  and  the  Base  Line,  respec- 
^tivelyjuaejreference'ljnes.  Then  lines  run  parallel  to  these 
1  'at" intervals*  of«l six1  miles,  divide  the  territory  into  squares 
each  containing  36  square  miles.  In  this  region  any  piece  of 
land  is  easily  located  by  indicating  its  distances  by  squares 
from  these  two  reference  lines.  In  short,  since  our  knowl- 
edge is  practically  all  relative,  the  principles  of  Analytical 
Geometry  lie  at  the  foundation  of  all  our  accurate  thinking. 

ART.  2.  The  two  intersecting  lines  are  called  Co-ordinate 
Axes,  and  their  point  of  intersection  is  called  the  Origin. 

In  the  system  most  frequently  used,  the  axes  meet  at 
right  angles,  and  hence  it  is  known  as  the  rectangular 
system.  In  comparatively  rare  instances  it  is  desirable  to 
have  the  lines  oblique  to  each  other,  when  the  system  is 
known  as  oblique. 

ART.  3.  The  vertical  axis  is  called  the  axis  of  ordinates 
and  the  horizontal  axis,  the  axis  of  abscissas. 

ART.  4.  Distances  are  always  measured  from  either  axis, 
parallel  to  the  other;  hence  when  the  system  is  rectangular, 
the  distances  mean  always  perpendicular  distances.  The 
distance  of  any  point  from  the  axis  of  ordinates  (right  or 
left),  measured  parallel  to  the  axis  of  abscissas,  is  called 
the  abscissa  of  the  point,  usually  represented  by  x.  The 


Analytical  Geometry.  3 

distance  from  the  axis  of  abscissas  (up  or  down),  measured 
parallel  to  the  axis  of  ordinates,  is  called  the  ordinate  of  the 
point,  usually  represented  by  y. 

ART.  5.  Clearly  if  we  would  be  accurate  we  must  dis- 
tinguish between  distance  to  the  right  and  to  the  left,  and 
upward  and  downward.  For  instance,  suppose  it  is  required 
to  locate  a  point  whose  abscissa,  x  =  5  and  ordinate,  y  =  2 ; 
it  is  plain  that  the  point  might  be  located  in  any  one  of 
four  positions:  to  the  right  5  units  and  up  2  units;  to  the 
left  5  and  up  2;  to  the  right  5  and  down  2;  or  to  the  left 
5  and  down  2. 

If,  however,  it  is  agreed  that  abscissas  measured  to  the 
right  from  the  axis  of  ordinates  shall  be  called  plus,  and 
those  to  the  left,  minus;  and  that  ordinates  measured  up- 
ward from  the  axis  of  abscissas  shall  be  called  plus,  and 
those  downward,  minus,  there  need  be  no  confusion. 

x  =  +  5,  y  =  +2  will  then  indicate  definitely  the  first 
position  referred  to  above;  x  =  —  5,  y  =  +  2,  the  second; 
x=  +  5,y  =  -2  the  third,  and  x  =  -  5,  y=  —  2,  the  fourth. 

ART.  6.    The  intersecting  axes  evidently  divide  the  sur- 


Y' 
Fig.  2. 

rounding  space  into  four  parts  called  quadrants,  numbered 
i,  2,  3,4,  from  the  axis  of  abscissas  (usually  called  the  X-axis) 


4  Analytical  Geometry. 

around  to  the  left  back  to  the  X-axis  again.  Thus  XOY  is 
quadrant  i;  X'OY  is  quadrant  2;  X'OY'  is  quadrant  3 
(Fig.  2). 

ART.  7.  To  locate  a  point  let  it  be  required  to  locate  the 
point  x=  —  5,  y  =  +  3%  [written  for  brevity  (-5,  34)]. 
Let  the  axes  be  XOX'  and  YO#'  as  in  Fig.  3. 

By  what  has  been  said  the  point  is  located  5  units  to  the 
left  of  the  Y-axis  and  3^  units  above  the  X-axis. 

Since,  it  is  a  matter  of  relative  position  only,  any  con- 
venient unit  may  be  used,  if  it  is  maintained  to  the  end  of 
the  problem;  say  in  this  case  J". 

Then  measuring  5  units  or  f"  to  the  left  on  the  X-axis, 


and  from  there  3! 

' 
JP(-6,8» 

1     ,    ,    ,     , 

units  or  ^-   =  ^"  upward  parallel  to 

f                                     the    Y-axis    we 
locate  the  point  P 
as  in  Fig.  3. 
The  point  (o,  2) 
is  clearly  on  the 

0 
I 

(1*,0)                          Y-axis,    2   units 
above   the  origin, 
because    the    ab- 
scissa is  zero,  and 
'                                     since  the  abscissa 
ig'3'                                  is     the     distance 

from  the  Y-axis,  this  point  being  at  no  distance,  must  be  on 
the  Y-axis.  Likewise,  the  point  (ij,  o)  is  on  the  X-axis 
i  J  units  to  the  right. 

EXERCISE   I. 

Locate  the  following  points: 

i-    (3»    2)>     (~  2,  -i),     (1,  -  3i),     (o,  i),     (-2,    o), 
(o,  o)  (-  6,  5),  (a,  -  f). 


Analytical  Geometry.  * 

2.  The  points   (o,  2^),    (  -  3,     -2)  and   (ij,     -  2}) 
are  the  vertices  of  a  triangle.     Construct  it. 

3.  Construct  the  quadrilateral  whose  vertices  are 

(-  i,  2),  (3,  5),  (2,  -  3)  and  (-  2,    -  2). 

4.  An  equilateral  triangle  has  its  vertex  at  the  point 
(o,  4)  and  its  base  coincides  with  the  X-axis.     Find  the  co- 
ordinates of  its  other  vertices  and  the  length  of  its  sides. 

5.  The  two  extremities  of  a  line  are  at  the  points  (—3,  4) 
and  (5,  4).     What  is  its  position  relative  to  the  axes? 

6.  How  far  is  the  point  (—  3,  4)  from  the  origin? 

7.  The  extremities  of  a  line  are  at  the  points  (3,  5)  and 
(—  2,  i),  respectively.     Construct  it. 

8.  The  extremities  of  a  line  are  at  the  points  (—  3,  —5) 
and  (3,  5).     Show  that  it  is  bisected  at  the  origin. 

9.  By  similar  triangles  find  the  point  midway  between 
(-  2,  5)  and  (4,  -  i). 

10.  A  line   crosses  the  axes    at  the   points  (15,  o)  and 
(o,  8).     What  is  its  length  between  the  axes. 

THE   POLAR   SYSTEM. 

ART.  8.  Since  two  dimensions  are  sufficient  to  locate  a 
point  in  a  plane,  it  is  readily  possible  to  use  an  angle  and 
a  distance,  instead  of  two  distances. 

By  convention  the  angle  is  estimated  from  a  fixed  line 
around  counter-clockwise;  the  revolving  line,  called  the 
radius  vector,  is  pivoted  at  the  left  end  of  the  fixed  line, 
which  is  called  the  initial  line,  and  the  pivotal  point  is 
known  as  the  pole. 

The  angle  is  estimated  either  in  degrees,  minutes,  and 
seconds  or  in  radians. 

ART.  9.  A  radian  is  defined  as  the  central  angle  which 
is  measured  by  an  arc  equal  in  length  to  the  radius. 


6  Analytical  Geometry. 

Since  the  circumference  of  a  circle  is  equal  to  2717-  (where 
r  is  the  radius  and  TT=  3.1416)  and  also  contains  360°, 

27ZT  =    360° 


and  r  =  ^-  =  ^-  =  i  radian. 

27T  7T 

Hence  the  number  of  radians  in  any  angle 

*JL    JL 

"    180   "    180   7 

7T 

That  is,  the  number  of  radians  in  an  angle  is  the  same 
fraction  of  TT,  that  the  angle  is  of  180°. 
For  example 

60°  = •  7i  radians  =  —  TT  radians. 

180  3 

TT  radians  =  —  TT  radians. 


180  8 

22 <°  =  — 5.  TI  radians  =  -5  n  radians,  etc. 
180  4 

ART.  10.  It  is  agreed  for  the  sake  of  uniformity  that 
an  angle  described  by  the  radius  vector  from  its  original 
position  of  coincidence  with  the  initial  line,  counter-clock- 
wise, shall  be  positive;  in  the  contrary  direction,  nega- 
tive. 

That  when  the  distance  to  the  point  is  measured 
along  the  radius  vector  forward,  it  shall  be  positive; 
when  measured  on  the  radius  vector  produced  back- 
ward through  the  pole  it  shall  be  negative.  For  example, 

the   point  (3,  —  J  would  be  located  thus  (Fig.  4)  : 

Draw   an   indefinite   line   OB    (representing  the   radius 

vector)  making  an  angle  of  —  radians  =     -  of  180°  =  60° 

3  3 


A  nalytical  Geometry,  7 

with  the  fixed  initial  line  OA;  measure  off  3  units  on  the 
radius  vector  from  the  pole,  and  the  point  P  is  located  (see 
Fig.  4). 


If  the  point  had  been  (—  3,  -J  the  3  units  would  have 

been  measured  back  toward  B'  to  P'.      If  the  angle  had 

been  —  —  the  radius  vector  would  have  taken  the  positive 

3 
direction  OB". 

The  usual  notation  for  co-ordinates  in  the  polar  system 
is  (r,  0)  or  (p,  6}. 

EXERCISE   II. 
i.    Locate  the  points: 


-  (24,75°),    (-4,  -30°). 


8  A  nalytical  Geometry. 

2.  Express  the  following  radians  in  degrees: 

7T  T.7T        71         ZTC         77T 

?    r'S'    7'    8'     8  •     16'   2'4" 

3.  Express  in  radians: 

35°,  40°,  45°,  6yi°,  75°,  150°,  120°,  -225°,  -  195°. 

4.  Construct  the  triangle  whose  vertices  are, 


5.    Construct  the  quadrilateral  whose  vertices  are, 


What  kind  of  quadrilateral  is  it  ? 

6.  The  extremities  of  a  line  are  the  points  (6,   —  ]  and 

(—6,  —  -  —  j.     How  is  the  line  situated  with  reference  to 

the  initial  line? 

7.  Construct  the  equilateral  triangle  whose  base  coin- 
cides with  the  initial  line  and  whose  vertex  is  the  point 


8.   The  co-ordinates  of  a  point  are  (5,  — ).    Give  three 

V      47 
other  wrays  of  denoting  the  same  point. 

AREA  OF    A    TRIANGLE. 

ART.  ii.  The  system  of  rectangular  co-ordinates  affords  a 
ready  method  of  expressing  the  area  of  any  triangle  when 
the  co-ordinates  of  its  vertices  are  known. 


Analytical  Geometry.  9 

Let  ABC  (Fig.  5)  be  any  triangle.  Draw  the  perpen- 
diculars AD,  BE  and  CF  from  the  vertices  to  the  ^-axis. 
Then  the  co-ordinates  of  A  =  (OD,  AD);  of 


Fig.  5. 

B  =  (OE,  BE);  of  C  =  (OF,  CF);  say,(-  *',  /),  (*",  f) 
and  (*"',  /"). 

Now  the  figure  ABCFD  is  made  up  of  the  trapezoids 
ABED,  and  BCFE;  and  if  from  ABCFD  we  take  ACFD 
the  triangle  ABC  remains,  that  is, 


ABED  +  BCFE  -  ACFD  =  ABC. 


(a) 


By  geometry,  area  ABED  =  J  (AD  +  BE)   DE. 

But  AD  =  /,  BE  =  /',  and  DE  =  DO  +  OE=  -x' 

3*. 

.'.  area  ABED  =  J  (/  +  /')  (x"  -  #')• 

Also  area  BCFE  =  J  (BE  +  CF)  EF. 

But  BE  =  V',  CF=//r  and  EF  =  OF-OE  =x*'-x*. 

.'.  area  BCFE  =  J  (/  +  f'}  (*?'  -  **)• 

Again;  area  ACFD  =  J  (AD  +  CF)  DF. 

But  AD  =  /,  CF=y"andDF=DO  +  OF  =  - 

.'.  area  ACFD  =  (/  +  /")  (^  -  ^). 


10 


Analytical  Geometry. 


Substituting  in  (a): 

Area  ABC  =  J  (/  +  /')  (x*  -  *')  + 

4  (/  +  /')  (*"'  ~  *")  -  i  (/  +  /")  (*"'  -  *0  = 

j  [yy  -  x'f  +  *"'/'  -  off  +  x'f-xnf  /]. 

The  symmetrical  arrangement  of  the  accents  in  this 
Expression  is  manifest. 

Example:  Find  the  area  of  the  triangle  whose  vertices  are 
(2,  3),  (-  i,  4)  and  (3,  -  6).  Let  (2,  3)  be  (*',  /); 
(-  i,  4)  be  (*",  /'),  and  (3,  -  6)  be  (*",  /").  Then 
area  =  i  [(-  i  X  3)  -  (2  X  4)  +  (3  X  4)  -  (- 1  X  -6) 
+  (2  X  -  6)  -  (3  X  3)]=  if-  3  -  8  +12-6-12-9] 

=  -  13- 

The  minus  sign  has  no  significance  except  to  indicate 
the  relation  of  the  trapezoids. 


Fig.  6. 

Polar  System  :  A  reference  to  Fig.  6  will  show  that  a 
similar  process  will  give  the  area  of  ABC,  when  its  vertices 
are  given  in  polar  co-ordinates. 

For    area    ABC  =  ABO  +  OBC  -  OAC. 

Area    ABO  =  i  AO  X  OB  sin  AOB. 

AO  -  /,  OB  =  r»  and  AOB  =  (Of  -  0"}. 

A  similar  treatment  of  OBC  and  AOC  will  give  the  areas 
of  all  the  triangles. 


CHAPTER   II. 
LOCI. 

ART.  12.  Whenever  the  relation  between  the  abscissa 
and  ordinate  of  every  point  on  a  line  is  the  same,  the  expres- 
sion of  this  relation  in  the  form  of  an  equation  is  said  to 
give  the  equation  of  the  line.  For  example,  if  the  ordinate 
is  always  4  times  the  abscissa  for  every  point  on  a  line, 
y  =  4  x  is  called  the  equation  of  the  line. 

Again,  if  3  times  the  abscissa  is  equal  to  5  times  the 
ordinate  plus  2,  for  every  point  on  a  line,  then  3  x  =  57+  2 
is  the  line's  equation. 

ART.  13.  Clearly  since  an  equation  represents  the  rela- 
tion between  the  abscissa  x  and  the  ordinate  y  for  every 
point  on  a  line,  if  either  co-ordinate  is  known  for  any  point 
on  the  line,  the  other  one  may  be  found  by  substituting 
the  known  one  in  the  equation  and  solving  it  for  the 
unknown. 

For  example,  let  2  y  =  j  x  —  i  be  the  equation  for  a 
line,  and  a  point  is  known  to  have  the  abscissa,  x  =  2. 
To  find  its  ordinate,  substitute  x  =  2  in  the  equation; 
2  y  =  7  (2)  -  i  =  14  —  i  =  13;  y  =  6J.  Therefore  the 
ordinate  corresponding  to  the  abscissa,  x  =  2,  is  6  J. 

Further,  if  the  equation  is  given,  the  whole  line  may  be 
reproduced  by  locating  its  points.  If  x  for  example  be 
given  a  series  of  values  from  o  to  10  inclusive,  by  substi- 
tuting these  values  in  the  equation,  the  corresponding 
values  of  y  are  found,  and  n  points  are  thus  located  on 
the  desired  line.  If  more  points  are  needed  the  range  of 

ii 


12  Analytical  Geometry. 

values  for  x  may  be  indefinitely  extended,  and  if  these 
points  are  joined,  we  have  the  line.  For  example,  let  the 
equation  of  a  line  be  x2  +  y2  =  9,  to  reproduce  the  curve 
represented.  For  convenience  in  calculating  solve  for  y\ 


Then  give  x  a  series  of  values  to  locate  points  on  this  line. 


_~_4  =  ± 
±  \/9  —    9  =  ± 


The  last  value  for  y  shows  that  the  point  whose  abscissa 
is  4  is  not  on  the  curve  at  all;  and  since  any  larger  values 
of  x  would  continue  to  give  imaginary  values  for  y,  the 
curve  does  not  extend  beyond  x  =  3. 

Since  we  have  given  x  only  positive  values  so  far,  all 
our  points  so  determined  lie  to  the  right  of  the  Y-axis. 
To  make  the  examination  complete,  let  x  take  a  series  of 
negative  value  thus: 

If  x  =  -  i  y  =  ±  Vg  -  i  =  ±  VjF=  ±  2.83. 
If  #  =  —  2  y=±  "N/9  —  4  =  ±  \/5  =  ±  2.24. 
If  #  =  —  3  v=±  V^  —9=0=  V  o. 

The  similarity  of  these  results  shows  that  the  curve  is 
symmetrical  with  respect  to  the  axes,  that  is,  it  is  alike 
on  both  sides  of  the  axes. 

If  now  these  points  are  located  with  respect  to  the  axes 
XX'  and  YY'  and  are  joined,  the  result  is  an  approxima- 
tion to  the  curve;  it  is  only  an  approximation  because  the 
points  are  few  and  not  close  enough  together. 

The  result  is  shown  in  Fig.  7,  using  J  inch  as  a  unit  for 


Analytical  Geometry, 


scale.     The   points  are  (o,  +3),   (o,    -  3)   [being  A  and 
A'  in  the  figure],  (i,  \/8),  (i,  -  \/8)  [being  B  and  B'], 
(2,  \/7),  (2,  -  \/J)  [being  C  and  C'],  (3,  o)  [G],  _ 
( -  i,  \/8),  (-  i,  -  \/8)  [D  and  D']  ( -  2,  \/S) 
(-  2,  -  \/5)  [E  and  E'J  and  (-  3,  o)  [F]. 


A' 


Fig.  7. 

Clearly  if  more  points  are  needed  to  trace  the  curve 
accurately  through  them  (as  is  the  case  here),  it  is  neces- 
sary to  take  more  values  of  x  between  —3  and  +3,  for 
example : 

#=o     ;y  =  ±  \/9  =  ±  3. 
x  =  .2     y  =  ±  \/9  -  .04 

y  =  ± 


:  V8.90  =  ±  2.99. 
-  .16=  ±  \/8.84=  ±  2.97. 
x  =  .6  ?  =  ±  \/9  -  .36  =  ±  \/8.64  =  ±  2.94. 
*  =  .8  y  =  ±  \/9  _  .64  =  ±  \/8^6~=  ±  2.89. 
^=  i  ^=±^9-  i=±  Vs"  =  ±  2.83,  etc. 

Making  a  similar  table  for  the  corresponding  negative 
values  of  x,  the  result  is  three  times  as  many  points  on  the 


Analytical  Geometry. 


curve  as  before,  and  as  they  are  closer  together  the  curve 
is  much  more  readily  drawn  through  them,  and  it  will  be 
much  more  accurate. 

Take  another  example :  9  x2  +  16 
Solving  for  v;  y  =  ±  f  V 16  —  x2 
Then  if 

y=  ±  |Vi6=  ±  3. 
y  =  ±  fvi6  —  .04= 
y  —  ± 
y  =  ± 

y=  ± 


=  O 

=  ±  .2 

=  ±  .4 

=  ±  .6 

=  ±  .8 

=  ±  i 


=  144. 


=  ±2.99. 
.16  =  ±  f  ^15.84  =  ±  2.98+ 

-  .36=  ±  1^15.64  =  ±  2.96. 

± 


-  .64 


=  ±  2.94. 


y==  ±  |Vi6  -  i   ==  ±  fVis      =±2.9,etc. 
The  result  is  indicated  in  Fig.  8,  same  scale  as  before. 


Fig.  8. 

ART.  14.  Clearly  a  curve  can  be  traced  thus  represent- 
ing almost  any  form  of  equation. 

Suppose  the  equation  of  ~  y#2+  7#+  15  =  ^  is  given. 
The  location  of  a  number  of  points  by  giving  x  a  series  of 
values  and  calculating  corresponding  values  of  y  from  the 
equation,  will  enable  us  to  draw  through  them  the  curve 
represented  by  the  equation.  In  most  cases,  there  will  be 
certain  values  of  x  which  will  make  the  value  of  y  zero; 


Analytical  Geometry.  15 

such  values  of  x  will  be  roots  of  the  equation  x3  —  7  xz 
+  7  #  +  15  =  o,  that  is,  these  values  of  x  indentically 
satisfy  this  equation. 

But  if  y  is  zero  for  a  point,  the  point  must  be  on  the 
X-axis,  for  by  definition  the  value  of  y  is  the  distance 
from  the  X-axis  to  the  point,  hence  the  curve  must  cross 
the  X-axis  at  those  points  where  y  is  zero.  If  then  none 
of  the  values  given  to  x  make  y  exactly  zero,  but  do  make 
y  change  from  a  positive  value  for  one  value  of  x  to  a 
negative  value  for  the  next,  or  vice  versa,  it  must  pass 
through  zero  to  change  from  one  sign  to  the  other,  and 
hence  the  curve  must  cross  the  X-axis. 

As  an  illustration,  take  the  equation  x3  —  5  x2  +  x 
-\-  ii  =  y.  As  before  make  a  table  of  values  of  x  and  y, 
and  locate  the  points  as  follows: 

If 


X  = 

o 

y  = 

ii. 

X  = 
X  = 

•5 

i 

yy  = 

10.375. 

8. 

X  = 

i-5 

y*= 

4.625. 

X  = 

2 

y  = 

i. 

x  — 

2-5 

y  = 

-  2.125. 

X  = 
X  = 

3 
3-5 

y  = 

-  4- 
-  3.875. 

x  — 

4 

y  = 

—  i. 

X  = 

4-5 

y  = 

5-375- 

X  = 

-  i 

y  = 

4- 

*=  -  1.5  y=  -  5-125- 

The  curve  connecting  these  points  crosses  the  X-axis  at 
three  points;  one  between  2  and  2.5;  one  between  4  and 
4.5,  and  one  between  --  i,  and  --  1.5.  Hence  the  three 
roots  of  the  equation  x3  —  5  x  2  +  x  +  1 1  =  o  are  be- 
tween 2  and  2.5;  between  4  and  4.5,  and  between  —  i  and 


1 6  Analytical  Geometry. 

If  the  values  of  x  in  the  above  table  had  been  taken 
closer  together,  the  points  of  crossing  would  have  been 
more  accurately  known. 

INTERSECTIONS. 

ART.  15.  The  point  (or  points)  in  which  two  lines 
intersect,  being  common  to  both  lines,  its  co-ordinates 
must  satisfy  both  equations,  that  is,  the  equations  of  the 
two  lines  are  simultaneous  for  this  point  (or  these  points) 
and  hence  if  the  equations  be  solved  as  simultaneous  by 
any  of  the  processes  explained  in  algebra,  the  resulting 
values  of  x  and  y  will  be  the  co-ordinates  of  the  point  (or 
points)  of  intersection.  For  example  : 

To  find  the  points  of  intersection  of  the  circle  x2  -f  y2  = 
24  and  the  parabola  y2  =  10  x.     By  substitution   of  the 
value  of  y2  from  the  parabola  equation  in  the  circle  equa- 
tion, 

x2  +  10  x  =  24  x2   +  10  x  +  25  =  49. 

x   +    5  =  ±  7  x     =     2,     or    --  12 

y=  ±V/2o,     or,    ±V  —  120. 

The  second  pair  of  values  for  y  being  imaginary  shows 
there  are  but  two  real  points  of  intersection,  (2,  +  V2o) 
and  (2,  —  \/2o).  Verify  by  construction. 

EXERCISE    III. 
Loci  with  Rectangular  Co-ordinates. 

1.  Express  the   equation  of  the  line  for  every  point  of 
which  the  ordinate  is  f  of  its  abscissa. 

2.  Express  the  equation  of  the  line  for  every  point  of 
which  f  the  abscissa  equals  {  of  the  ordinate  +  i. 


A  nalytical  Geometry.  1 7 

3.  Express  the  equation  of  the  line,  for  every  point  of 
which  9  times  the  square  of  its  abscissa  plus  16  times  the 
square  of  its  ordinate  equals  144. 

4.  Construct  the  locus  of  x2  =  8  y. 

5.  Construct  the  locus  of  (x  —  2)2  +  y2  =  36. 

6.  Construct  the  locus  of  xy  =  16. 

7.  Construct  the  locus  of  x2  -f-  4  y2  =  4. 

8.  Construct  the  locus  of  25  x2  —  36  y2  =  900. 

9.  Construct  the  locus  of  3  x  —  2  y  =  5. 

10.  Construct  the  locus  of  %  x  —  f  =  —  y. 

11.  Construct  the  locus  of  x  =  7. 

12.  Construct  the  locus  of  y  =  —  5. 
Find  the  points  of  intersection  of: 

13.  (x-  i)2  +  (y  -  2)2=  16  and  zy-  x=  3. 

14.  2x  —  $y=  7  and  i  x  +  y  =  f . 

15.  x2  +  y  =  9  and  #2  =  8  ;y. 

16.  #2  +  y2  =  16  and  2  x2  +  3  y2  =  6. 

17.  x2  +  y2  =  25  and  4  y  =  3  #  +  25. 

18.  Find  the  vertices  of  the  triangle  whose  sides  are 

x  —  y  =  i. 

2^  +  ^=5  and  3^-2*=  7. 

ART.  1 6.  If  the  equation  of  a  locus  is  expressed  in  polar 
co-ordinates,  the  method  of  procedure  is  exactly  similar 
to  the  cases  already  discussed. 

The  presence  of  trigonometric  functions  introduces  no 
difficulties.  For  example:  To  construct  the  locus  of 
r  =  4(1  —  cos  6).  Give  0  a  series  of  values,  and  com- 
puting r  for  each,  as  follows: 

If  0  =  o,         r  =  o  since  cos  o  =  i. 
If  0  =  5o?       r  =  4  (z  _  .996)  =    .OI6. 

If  6  =  10°,     r  =  4  (i  -  .98)    =    .08. 
If  0=  15°,     r*4(i-  -97)    =    -12. 


i8 


Analytical  Geometry. 


If  0  =  20°, 

if  e  =  3o°, 

If  6  =  40°, 

if  e  =  50°, 

if  e  =  60°, 


r  =  4  (i  -  .94)  =    .24. 

r=  4  (i  -  -87)  :        .52- 

r=  4  (i  -  .77)  =     .92. 

r  =  4  (i  —  .64)  =  1.44. 

r  =  4  (i  -  -5  )  =  2.     ,etc. 


Fig.  9. 


Completing  the  table  to  0 
curve  as  in  Fig.  9. 


360°  and  plotting  we  get  a 


TRANSCENDENTAL    LOCI. 

ART.  17.  Certain  curves  have  what  are  known  as  trans- 
cendental equations,  that  is,  equations  which  cannot  be 
solved  alone  by  the  algebraic  processes  of  addition,  sub- 
traction, multiplication,  and  division. 


Analytical  Geometry.  19 

For  example,  y  =  log  x. 

The  loci  of  such  equations  are  found  in  the  usual  way, 
by  giving  to  one  of  the  co-ordinates  a  series  of  values  and 
finding  corresponding  values  for  the  other  from  tables. 

EXERCISE   IV. 

1.  Find  the  locus  of  r2  =  9  cos  2  0. 

2.  Find  the  locus  of  r  =  10  cos  0. 

3.  Find  the  locus  of  r  = —  -  • 

i  —  cos  0 

4.  Find  the  locus  of  r  = . 

5  +  3  cos  6 

5.  Construct  y  —  sin  x. 

6.  Construct  x  =  log  y. 


MISCELLANEOUS    CURVES. 

ART.  1 8.  Curve-plotting  is  very  widely  applied  in  all 
modern  scientific  research,  to  represent  graphically  the 
results  of  observation.  This  method  of  presentation  has 
the  immense  advantage  of  showing  at  a  glance  the  com- 
plete result  of  an  investigation. 

For  example,  if  a  test  is  made  of  the  speed  of  an  engine 
relative  to  its  steam  pressure,  the  pressures  being  repre- 
sented as  abscissas  (by  x)  and  the  corresponding  speeds  as 
ordinates  (by  y},  a  smooth  curve  drawn  through  the  points 
determined  by  these  co-ordinates  will  reveal  at  once  the 
behavior  of  the  engine.  Especially  does  this  method  aid 
in  comparisons  of  different  series  of  observations  of  the 
same  kind. 


20 


Analytical  Geometry. 


Suppose,    for   example,  it    is    desired   to    represent  thus 
graphically  the  course  of  a  case  of  fever. 
The  observations  are  as  follows:  — 


7  A.M. 

temperature  100 

8  A.M. 

"     i  oof 

9  A.M. 

"    IOII 

IO  A.M. 

I02f 

II  A.M. 

''      103 

12  M. 

io3f 

I  P.M. 

103 

2  P.M. 

I02f 

3  P.M. 

101 

Regarding  the  time  of  taking  observations  as  abscissas 
and  the  temperatures  as  ordinates,  using  any  desired  scale, 
the  result  may  be  represented  as  follows,  in  Fig.  10. 


104" 
103° 
102° 
101° 

100° 

Q8° 

^ 

^ 

J 

^ 

s 

/ 

,/ 

\ 

' 

r 

1 

- 

IM 

:   i 

INE 

7      8      9      IO     II      12      1 

Fig.  10. 
Fig.   10. 


234 


The  figure  shows  at  a  glance  that  the  maximum  was  at 
noon. 


Analytical   Geometry.  21 

•  Again;  in  the  test  of  an  I-beam  the  following  observations 
were  taken. 

TEST    OF    CAST-IRON. 
Stress  Pounds.         Unit  Elongation* 

O  O 

6,950  4.97 

12,940  11.44 

6,110  6.06 

o  1.12    (permanent  set) 

4,640  4.16 

8,780  7.63 

12,300          10.78 
15,420  15.2 

11,900  12.38 

8,370  9.42 

4,960  6.66 

113  2.41 

Plot  the  curve. 

EXERCISE   V. 


CHAPTER   III. 
THE    STRAIGHT    LINE. 

ART.  19.  Since  two  points  determine  a  straight  line 
and  two  points  imply  two  conditions,  there  will  be  in  the 
equation  to  a  straight  line,  two  fixed  quantities  (called 
constants),  which  must  be  predetermined  for  every  straight 


Fig.  ii. 

line.     These  constants  may  be  furnished  by  two  fixed  points, 
or  by  a  point  and  an  angle,  evidently. 

To  determine  the  equation  of  a  given  straight  line,  then, 
it  is  necessary  to  express  the  relation  between  the  co-ordi- 
nates of  any  (that  is,  every)  point  on  the  line,  in  terms  of 
the  two  given  constants. 

22 


Analytical  Geometry.  23 

Suppose  first  we  take  a  point  on  the  y-axis,  through 
which  the  line  must  pass,  and  determine  its  posi- 
tion by  giving  its  distance  from  the  origin  measured  on 
this  axis. 

Call  this  distance,  b;  and  say  the  line  makes  an  angle  a 
with  the  #-axis;  the  angle  to  be  estimated  as  in  trigo- 
nometry, positively,  that  is,  counter-clockwise,  from  the 
tf-axis.* 

It  is  required,  then,  to  determine  the  relation  between 
the  co-ordinates  of  any  point  P,  selected  at  random,  on  the 
line  AB  (Fig.  IT),  using  b  and  any  convenient  function  of  a. 

Drawing  thej_  PR,  OR   =  abscissa  of  P  =  x, 

PR  =  ordinate  of  P  =  y,  OS  =  b. 
Z  BTR  =  a, 

The  character  of  the  figure  would  suggest  the  use  of  the 
similar  triangles  TSO  and  TPR,  but  a  simple  observation 
shows  that  only  the  sides  b  and  y  are  known;  on  the  other 
hand  we  know  the  angle  a,  and  a  line  through  S  ||  to  the 
#-axis,  from  S  to  PR,  will  be  equal  in  length  to  OR  and 
will  also  make  the  angle  a  with  AB  (alternate  angles  of 
parallel  lines). 

Call  this  line  SN.  Then  in  the  triangle  SPN,  Z  PSN 
=  a  SN  =  OR  =  xt  and  PN  =  PR  -  NR  =  PR  -  SO 
=  y  —  b.  PN  and  SN  being  respectively  opposite  and 
adjacent  to  a  in  the  right  triangle  SPN,  we  have, 

PN          -  b 


*  The  conventions  as  to  positive  and  negative  direction  for  lines, 
and  positive  and  negative  revolution  for  angles,  is  maintained  in 
Analytical  Geometry,  as  indeed  is  necessary  in  order  to  accomplish 
consistent  results. 


Analytical  Geometry. 


Let  tan  a  be  represented  by  m; 

-  b 


khen 


m  — 


mx  =  y  —  b 

y  =  mx  +  b (A) 

which  expresses  the  relation  between  the  co-ordinates  of 
of  any  point,  P,  and  hence  of  every  point  on  the  line  in 
terms  of  the  known  constants  m  and  b.  .'.  y  =  mx  +  b 


K 


Fig.    12. 

is  the  equation  of  AB.  Had  the  line  crossed  the  first  quad- 
rant the  construction  would  have  been  as  in  Fig.  12  and 
we  would  have 

tanPSN=    ^?, 

SN 


or  tan  (180  —  a)  = 
—  tan  a  = 


b  - 

x 


X 

X 

y  =  mx  +  b  as  before. 


Analytical  Geometry.  25 

m  is  called  the  slope  of  the  line  and  b  its  ^-intercept.    The 
equation  is  called  the  slope  equation  of  a  line. 

If  m  =  o  in  the  equation  to  a  straight  line,  then  it  takes 
the  form  y  =  b,  which  is  plainly  (since  if  m  =  o,  a  =  o) 
a  line  ||  to  the  ac-axis.  If  b  =  o,  the  equation  becomes 
y  =  mx,  which  is  the  equation  of  a  line  through  the  origin, 
making  an  angle  whose  tangent  is  m  with  the  #-axis,  etc. 
Since  a'  may  be  either  acute  or  obtuse  depending  upon 
whether  the  line  crosses  the  2d  or  4th,  or  the  ist  or  3d 
quadrants;  and  b  may  be  either  plus  or  minus  depending 
upon  the  position  of  the  point  of  intersection  with  ^-axis, 
above  or  below  the  origin,  the  form, 


y=  —  mx  +  b  represents  a  line  crossing  quad.  I, 
=        mx  +  b  represents  a  line  across  quad.  II, 
=  —  mx  —  b  represents  a  line  across  quad.  Ill, 
c  —  b  represents  a  line  across  quad.  IV. 


y  = 
y 

y  =       mx 


ART.  20.  //  the  line  be  determined  by  two  points  (x',  /) 
and  (x",  /') ;  to  find  its  equation. 

Let  AB  (Fig.  13)  be  the  line,  P  and  Q  the  points  (V,  /) 
and  (x",  y"),  respectively. 

Take  any  point  P'  whose  co-ordinates  are  (x,  y).  Draw 
QR,  P'S  and  PT  _L  to  the  *-axis,  also  PL  J_  to  QR,  as 
it  is  clearly  here  a  case  for  similar  triangles. 

Then  in  the  similar  triangles  PLQ  and  PKP', 


P'K:KP::QL:LP,     or 


But      P'K  =  P'S  -  KS  =  P'S  -  PT  =  y  -  / 
KP  =  HP  -  HK  =  x'     -  x, 
QL  =  QR  -  LR  =  QR  -  PT  =  f  -  /, 


26  Analytical  Geometry. 

and          LP   =  LH  +  HP  =-**•+ 

y  —  y  _     y  —  y 


or  symmetrically, 


X  —  X* 


Xf 


(changing  sign  of  both)  which  gives  an  equation  between 
xt  y,  ocfy  y,  xf> ',  and  y  as  required. 

B\    (x",y"\ 

\  '/-v       >a      I 


L 

\ 

H 

y 

\ 

K 

.-\?(«'.^ 

S          T 


Fig.  13. 

The  same  result  might  be  reached  by  a  purely  analytical 
method  having  the  slope  equation  of  a  line  given. 

Let  the  slope  equation  of  the  line  AB  be  y  =  mx  +  b. 

Since  it  must  pass  through  the  points  P,  P'  and  Q,  the 
co-ordinates  of  these  points  must  satisfy  the  equation  of 


Analytical  Geometry.  27 

the  line,  since  the  equation  must  give  the  relation  between 
the  co-ordinates  of  every  point  on  the  line. 

Hence,  substituting  these  co-ordinates  successively  in 
the  equation  y  =  mx  +  b,  we  know  that  the  three  following 
equations  must  be  true,  if  P,  P'  and  Q  are  on  the  line: 

/  =  mx'  +  b     ......     (i) 

y   =  mx    +  b     ......     (2) 

f=mx"  +  b     ......     (3) 

But  since  the  line  is  to  be  determined  only  by  the  two 

points  P  and  Q,  neither  m  nor  b  are  known,  and  hence 

must  be  eliminated. 

Subtracting  (i)  from  (2)  and  (i)  from  (3),  we  get 

y    -  y  =  m  (x    -  x')       .     .     .     (4) 
and  f-y^m  (x"  -  *')       ...     (5) 


divide  (4)  by  (5); 


For  example:   Find   the  equation   of  the   line   through 
(-  2,  3)  and  (-4,  -  6). 

Let     (*',  /)  be  (-  2,  3)  and  (*",  /')  be  (4,  -  6)*. 
Substituting  in  (B), 


*  Since  (B)  is  perfectly  symmetrical  it  is  a  matter  of  indifference 
which  point  be  called  (#',  y')  and  which,  (x",  y").  The  results  are 
the  same.  It  is  to  be  observed  that  x  and  y  with  accent  marks 
usually  mean  definite  points,  while  general  co-ordinates  are  repre- 
sented by  unaccented  x  and  y.  So  that  substitutions  are  always 
made  for  the  accented  variables,  when  definite  points  are  involved. 


28 


Analytical  Geometry. 


ART  21.  When  the  line  is  determined  by  an  angle  and  a 
point  situated  otherwise  than  on  the  y-axis. 

Let  the  tangent  of  the  angle  be  m  and  the  point  be  (V,  y'). 
Then  y  =  mx  +  b  (i)  can  represent  the  slope  equation  to 
the  line.  This  equation  satisfies  the  condition  that  "the 
line  should  have  the  slope  m,  but  it  must  also  pass  through 
the  point  (V,  /). 

Hence,  if  y  =  mx  +  b  is  to  completely  represent  the 
line,  equation  yf  =  mx'  +  b  (2)  must  be  true. 

Since  b  is  a  third  and  unnecessary  condition,  it  must  be 
eliminated  between  (i)  and  (2). 

y  =  mx  +  b 

=>»*'  +  (Q* 


Subtract  (2)  from  (i); 


y  —yf  —  mx—  mx*  =  m  (x  —x'} 


Q 


Fig.  14. 

ART.  22.     When  the  line  is  determined  by  two  points,  one 
on  each  axis. 

*  It   is    to    be   observed    that   the    slope    equation    is    a    special 
form  of  (£)  where  (V,  y')  is   (0,  6). 


Analytical  Geometry. 


29 


Let  the  points  P  and  Q,  respectively  (0,  b)  and  (a,  0), 
be  the  determining  points  (Fig.  14),  and  let  y  =  mx  -\-  b 
be  the  slope  equation  of  the  line  AB;  then  b  =  b  and 
m  =  tan  PQX=  -  tan  PQO.  Also 

tanPQO  =  -•        .'.  m  =  -  -  . 
a  a 

Substituting  these  values  of  m  and  b  thus  expressed,  by 
a  and  b  in  the  slope  equation, 


V  =  —  —  x  +  b,     or      . 
a  b 


2-, 


(D)= 


[dividing  by  b  and  transposing]. 

This  form  is  known  as  the  intercept  equation  of  a  straight 
line,  since  a  and  b  are  called  the  intercepts  of  the  line  AB 
on  the  co-ordinate  axes. 

ART.  23.     There  is  still  another  form  of  equation  to  the 


Fig.   15- 

straight  line  determined   by  a  perpendicular  to  the  line 
*  The  same   result  could  be  derived  from    (B)   by  substituting 
(a,  o)  for  (V,  /)  and  (b,  o)  for  (x",  y"}. 


30  Analytical  Geometry. 

from  the  origin,  and  the  angle  which  this  perpendicular 
makes  with  the  #-axis. 

Let  OD  be  a  _L  to  the  line  AB  from  the  origin,  and  ft 
the  angle  it  makes  with  the  #-axis.  Let  P  (x,  y)  be  any 
point  on  the  line. 

Drawing  the  ordinate  (PE)  of  P,  we  have  two  similar 
right  triangles  ODF  (F  being  the  point  where  AB  crosses 
the  jc-axis)  and  PEF. 

Then  PE  :  OD  :  :  EF  :  DF  [homologous  sides]. 

Call  OD,  p,  and  OF,  a,  then  above  proportion  becomes 
y  :  p  :  :  (a-  x)  :  DF. 

But  in  the  right  triangle  ODF, 


;COS  ft  I  COS  ft 

*mP  =  p  I — £-r  —  x\     [extremes  and  means] 
s  ft  \cos  ft         I 

or  ysmft  =  —t—  -  x    [dividing  bv  p] 

cos  ft         cos  /? 

that  is,  y  sin  /?  +  #  cos  ft  =  p (E) 

This  is  called  the  normal  equation,  p  being  known  as  a 
normal. 

The  line  AB  is  plainly  a  tangent  to  a  circle  with  O  as  a 
centre  and  p  as  a  radius,  hence  we  are  practically  deter- 
mining the  line  AB  as  a  tangent  to  a  given  circle,  the  posi- 
tion of  the  radius  being  fixed  by  the  angle  ft. 

Exercise:  By  determining  the  values  of  a  and  b  from  the 

intercept  equation, h  —  =  i,  in  terms  of  p  and  ft,  derive 

a        b 

the  normal  equation  from  the  intercept  equation. 


A  nalytical  Geometry.  3 1 

ART.  24.  Each  equation  has  its  characteristic  form. 
For  instance,  the  slope  equation  y  =  mx  -f  b,  has  the 
form  of  a  first  degree  equation  solved  for  y,  hence  if  any 
first  degree  equation  be  solved  for  y,  it  may  be  compared 
directly  with  this  slope  equation.  For  example,  given  the 
equation  2  y  —  3  x  =  8.  Solving  for  y,  y  =  %  x  +  45  com- 
paring this  with  the  typical  form;  m  =  f  and  b  =  4. 

Hence  the  locus  of  2^  —  3^=8  may  be  constructed 
as  follows,  remembering  the  meaning  of  m  and  b,  (Fig.  16). 

First  to  construct  any  line  making  an  angle  whose  tan- 
gent is  f  with  the  ^-axis.  By  trigonometry  if  we  lay  off 
on  the  ;y-axis  a  distance  3  and  on  the  ^-axis  a  distance  2 


Fig.    16. 

(remembering  that  the  angle  must  be  measured  from  right 
to  left),  the  line  DE,  drawn  through  the  points  so  deter- 
mined makes  an  angle  whose  tangent  is  f  with  the  jc- 


for  tan.  FLO  =     ^  =  f>  nence  an7  line  drawn  ||  to  ED 
makes  the  same  angle.     If  this  line  is  drawn  through  the 


Analytical  Geometry. 


point  G,  4  units  above  the  origin  (b  =  4),  it  will  be  the 
required  line,  as  AB  in  the  figure. 

In  this  case  m  =  f  being  positive  shows  that  the  line 
crosses  either  the  2d  or  4th  quadrants,  and  b  =  4  being 
positive  shows  it  is  the  2d,  hence  the  construction. 

If  m  is  negative,  it  crosses  either  the  ist  or  3d  quadrants, 
and  the  sign  of  b  will  determine  which  one.  Hence  in  every 
case  we  know  where  to  make  the  construction  for  m. 

It  is  usually  easier  to  make  use  of  two  points  for  the  con- 
struction of  straight  lines,  and  these  points  are  most  easily 
determined  on  the  axes,  where  the  line  crosses  them. 

Since  the  equation  of  a  line  expresses  the  relation  between 
the  co-ordinates  of  every  point  on  the  line,  it  will  express 
the  relation  for  these  points  on  the  line  where  it  cuts  the 
axes;  but  at  these  points  either  x  or  y  is  o,  depending  on 
whether  it  is  the  y  or  the  #-axis.  Hence  to  find  the  inter- 


K/8/ 


Fig.  17. 

cept  on  the  #-axis,  set  y  =  o  in  the  equation  (for  at  the 
point  of  crossing  y  —  o);  the  value  of  x  will  then  be  the 
^-intercept.  Likewise,  to  find  the  y-intercept  set  x  =  o 
in  the  equation. 


Analytical  Geometry.  33 

In  the  preceding  example, 

2  y  —  3x=  8. 

Set  y  =  o,  0—3^=8 

x  =  —  §  (x  —  intercept). 

Set  x  =  o  2  y  —  0=8. 

y  =  4  (y  —  intercept). 

Hence  measuring  —  f  to  the  left  on  the  #-axis  and  4 
upward  on  the  ;y-axis,  the  line  passes  through  these  two 
points. 

ART.  25.  The  characteristic  property  of  the  intercept 
equation  is  that  the  right  hand  member  of  the  equation  is  i, 
and  the  other  member  consists  of  the  sum  of  two  fractions 
whose  numerators  are  respectively  x  and  y.  For  example, 
to  put  the  equation  3  x  —  4y  =  7  into  intercept  form. 
To  make  the  right  side  i,  the  equation  must  be  divided 
by  7. 

••-  f*  -  f  y=  i (O 

To  change  the  left  hand  side  to  the  sum  of  two  fractions 
having  x  and  y  only  for  numerators,  the  equation  may  be 
written  thus: 

x   ,      v 

j  +  "'- 

comparing  this  with  the  type  form, 

*+£-!, 

a        b 

evidently  a  =  J    and    b  =  —  |. 

These  values  may  be  verified  by  the  method  indicated  in 
the  last  article. 

Let  y  =  o  in  (i),  then  f  .r  —  o  =  i  x  =  %  =  a. 

Let  #  =  o,  then  o  —   ^=   i, 

7 

y  =      -    I  =    J. 

What  is  typical  of  the  normal  equation  ? 


34  Analytical  Geometry. 

ART.  26.  Any  equation  of  the  first  degree  in  two  vari- 
ables represents  a  straight  line. 

Any  equation  of  the  first  degree  in  two  variables  may 
be  represented  by 

A*  -f  Ey  =  C. 
This  equation  may  be  put  in  the  form 


which  is  clearly  the  slope  equation  of  a  straight  line,  whose 

A  C  A 

slope  is  —  -  and  y  -  intercept,  —  ;  that  is,  m  =  --  and 
B  B  B 

-f 

Again  :     The   equation   Ax  +  "By  =  C   may   be   put   in 

the  form    —  -f  21  =  i   (DJ  which   is  the  intercept  form, 
i_x        \-s 

A       E 

C  C 

where        _  and  —  are  the  two  intercepts. 
A          B 

Again:  To  put  Ax  +  By  =  C  in  the  normal  form, 
x  cos  ft  +  y  sin  /?  =  p,  it  is  necessary  to  express  cos  /?, 
sin  /?  and  p  in  terms  of  A,  B  and  C  (Fig.  18).  It  has 
been  shown  above  that  the  intercepts  OM  and  ON  (MN 

C  C 

being  the  line)  are  —  and  —  • 
B  A 

Since  Z  OMN  =  Z  PON  =  /?,  in  the  right  triangle 
MON, 

^ 

Sin  ft  =  -  ^  =  -  -, 

^>    A/A2  +  B2  VA2  +  B2 


Analytical  Geometry. 


35 


and  cos   ft  = 


VA2   +  B2 


In  the  similar  triangles  MON  and  PON,  OM  :  OP  :  : 
MN  :  ON, 


that  is, 


Whence 


'A2  +  B2) 

substituting  these  values  in  the  normal  equation, 
Ax         ,  Ey         _  C 

VA2  +  B2     • 


VA2  +  B2       VA2   +  B2 


(Et)* 


*  The  sign  of  \/A2  +  B2  is  readily  determined  from  the  sign  of 


C  in  Ax  4-  By  =  C,  for  p 


and  since  p  is  essentially 


, 

positive,  C  and  \/A2  +  B2  must  have  the  same  sign  that  this  equa- 
tion may  be  true. 


36  Analytical  Geometry. 

which  is  plainly  obtained  from  Ax  +  Ey  =  C,  by 
dividing  through  by  \/A2  +  B2,  that  is,  the  square  root 
of  the  sum  of  the  squares  of  the  coefficients  of  x  and  y. 
For  example,  to  put  3^  +  4^=  9  in  the  normal  form: 

In  this  case  \/A2  +  B2  =  Vf  +  42  =  ^25  =  5. 

Dividing  then  by  5;  3  x  +  4  v  =  9  becomes 


where  «*.  =  cos  /?,  —  =  sin  /?  and  —  =  p. 

5  5  5 

From  the  above  it  is  seen  that  a  general  equation  Ax  + 
B)/  =  C  can  assume  any  of  the  type  forms  for  a  straight 
line,  hence  it  may  always  represent  a  straight  line. 

ART.  26  (a).  Another  method  of  reducing  Ax  +  B_y  =C 
to  the  normal  form,  is  easily  derived  from  the  following 
consideration  : 

If  two  equations  both  represent  the  same  straight  line, 
they  cannot  be  independent  equations,  but  one  must  be 
obtained  from  the  other,  by  multiplying  it  through  by 
some  constant  factor,  like 

2  x  —  $y  =  i  and  8  x  —12  y  =4. 

That  is,  all  the  coefficients  in  one  are  the  same  number 
of   times   the   corresponding   coefficients   in   the   other,    as 
8=4X2,     12=4X3  and  4  =  4  X  i. 

Now  if  Ax  +  Bv  =  C  and  x  cos  /?  +  y  sin  /?  =  p  are 
to  represent  the  same  straight  line, 

then 
that  is, 


B  =  n  sin  /?    ..........     (2) 


Analytical  Geometry.  37 


To  find  n,  square  (i)  and  (2)  and  add; 
A2  =  n2  cos2  /? 
B2  -.  n2  sin2  /? 


= 

VA2 

+  B2 

= 

A 

[from 

V/A2 

+  B2 

B 

[from 

VA2 

+     lj  2 
X) 

C 

A2  +  B2  =  n2  (sin2  /?  +  cos2  /?)  =  n* 
[since  sin2  /?  +  c°s2  /?  =  i] 
or 

.'.  cos 
sin  a  = 


and  p  = 

VA2  +  B2 

For  sign  of  vA2  +  B2,  see  note  in  Art.  26. 

ART.  27.  From  what  was  said  about  intersections  under 
loci,  it  is  clear  that  if  two  equations  representing  straight 
lines  are  combined  as  simultaneous,  the  resulting  values 
of  x  and  y  are  the  co-ordinates  of  their  point  of  intersection. 
For  example: 

Let  2*- 3^=5 (i) 

•x  +  $y=  17 (2) 

be  the  equations  of  two  lines. 

Multiplying  (2)  by  2  and  subtracting; 

2  x  -    3  y  =  5 
2  x  +  io  y  =  34 
13?=.  29 

y=  29?  whence  x=  if. 

That  is,  these  two  lines  intersect  at  the  point  (£§,  f  f ). 
Verify  by  construction. 


38  Analytical  Geometry. 

EXERCISE    VI. 
Straight  Line. 

What  are  the  slope  and  intercepts  of  the  following  lines? 
Construct  them. 

i.   2y=3x  +  i.  2.   3^  +  2^  +  7=0. 

3-  sy=  -  x  -  6-          4.47-7*  +  i  =  o- 

5.   §*-iJ?=ii  6.    \y  -  2X  +  3  =y  +  i#- 

7-   oc  +  y=  o.  8.   y  =  -  3. 

9.  A  line  having  the  slope  f  cuts  the  ;y-axis  at  the  point 
(o,  —  3).     What  is  its  equation? 

10.  What  are  the  vertices  of  the  triangle  whose  sides  are 
2  y  —  x  +  i  =  o,  $y  +  x  =  2,  x=  —  2  y  +  i? 

11.  Find  the  vertices  of  the  quadrilateral  whose  sides  are 
x=y,  y+x=2,  $y  —  2x=$,  2x  +  y  =  -i. 

12.  The  vertices  of  a  triangle  are   (2,  o),    (—  3,   i), 
(—  5,  —4).     What  are  the  equations  of  its  sides? 

13.  A  line  passes  through  (—  3,  2)  and  makes  an  angle 
of  45°  with  the  #-axis.     What  is  its  equation  ? 

14.  What  is  the  equation  of  the  common  chord  of  the 
circles  (x  —  i)2  +  (y  —  3)2  =  50  and  x2  +  y2  =  25? 

15.  The  points  (6,  8)  and  (8,  4)  are  on  a  circle.     What 
is  the  equation  of  a  chord  joining  them  ? 

16.  Which   of    the   following   points    are    on    the    line 

2  y  =  ~ZOC+  2;    (2,    I),     (-2,    §),     (2,    -   2),    (5,2)? 

17.  What  is  the  slope  of  the  line  through  (i,  —  6)  and 

(-3,5)? 

1 8.  What  slope  must  a  line  with  the  ^-intercept   —  3 
have  that  it  may  pass  through  (—3,  2)  ? 

19.  Show  that   (i,  5)   lies   on    the  line  joining   (o,   2) 
and  (2,  8). 

20.  Show  that  the  line  joining  (—  i,  f)  and  (2,  —  3) 
passes  through  the  origin. 


Analytical  Geometry.  39 

ART.  28.  To  find  the  angle  between  two  intersecting  lines 
jrom  their  equations. 

Let  y  =  mx  +  b,  and  y  =  m'x  +  b',  be  the  equations  of 
two  intersecting  lines,  AB  and  CD,  in  Fig.  19. 


\ 


tan-^m 


. 


Fig.   19. 

Since  the  slopes  are  m  and  m'  respectively,  tan  FHX=  m 
and  tan  FGX=  w'. 

In  the  triangle  GFH,  formed  by  the  intersecting  lines  and 
the  x-axis,  the  external  angle 

FHX  =  HGF  +  GFH 
or  GFH  =  FHX- HGF (i) 

Call,  for  convenience,  GFH,   6;  FHX,  a;  and  HGF,  /9. 
Then  by  (i)   0=  a-  ? (ia) 

Since  the  result  must  be  expressed  in  m  and  w',  that  is, 
in  the  tangents  of  a  and  ft,  the  trigonometric  formula  for 


4o  Analytical  Geometry. 

the  tangent  of  the  difference  of  two  angles  (a  —  3)  must 
be  used,  that  is, 

ON         tan  a  —  tan  B          m  —  mf 
tan  (a  —  /?)  =  -  —  '-—  •  =  -          —  • 

i  +  tan  a  tan  p       i  +  ww' 

But  since     0  =  a  —  fi,  tan  0  =  tan  (a  -  /?). 

/.tan  0  =  ^^^-.    .  .    (F) 

i  +  mm' 

Which  enables  us  to  calculate  6  from  m  and  m'.  For 
example,  to  find  the  angles  between  the  two  lines 

f*-f  y=  i 
and 

i*  +  iy=  ij. 

Putting  these  equations  in  the  slope  form,  they  become, 

y=     S*  -  J 
y=  -  I*  +  !. 

Since  two  lines  intersecting  always  form  two  angles, 
which  are  supplementary  with  each  other,  and  since  the 
only  difference  that  can  result  in  the  formula 


tan  e  = 


i  +  mm 

from  interchanging  m  and  m'  is  a  reversal  of  sign,  that  is, 
a  change  from  the  value  of  0  to  its  supplement,  unless  it 
is  distinctly  specified,  that  the  angle  of  intersection  is  the 
acute  or  obtuse  angle,  it  makes  no  difference  which  slope 
be  called  m  or  mf . 

Say  in  above,         m  =  f  and  m'  =  —  f . 

Substituting  in  formula   (F), 

tan  Q  =        8  ~  f-i)      =  £±1=:  if  =  s|  =  4.9167. 

i  +  (f)  (-1)      i- I       i 

A  table  of  trigonometric  functions  will   show  from   this 
value  that  0  =  78°  -  30'  -  12"  +. 

Make  the  construction  and  test  with  protractor. 


Analytical  Geometry.  41 

ART.  29.     To    find    condition    for    perpendicularity    or 
parallelism  of  lines  from  their  equations. 
In  formula  (F), 


i  +  mm' 
When  the  lines  are_J_,  0  =  go0,  and  .*.  tan  $  =  oo  ;  that  is, 


i  +  mm' 

Since  a  fraction  whose  numerator  is  finite  equals  oo  only 
when  its  denominator  =  o,  .'.  in  this  case 

i  +  mm'  —  o  or  m'  =  —  —  .    .    .    .    .    (a) 

iff 

That  is,  two  lines  are  perpendicular  to  each  other  when 
their  slopes  are  negative  reciprocals. 

For  example,  3^—2^=5  and  2  x  +  3  y  =  n  are 
perpendiculars. 

When  the  lines  are  parallel,  0  =  o  and  hence, 
tan  0=o. 


rr-i    L  -  m  —  m 

That  is,        -      —  =o    or    m  —  m  .  —  o. 
T  +  mm 


Whence 


That  is,  their  slopes  are  equal.  These  conditions  enable 
us  to  readily  draw  a  perpendicular  or  a  parallel  to  a  given 
line  through  a  given  point. 

For  we  can  find  the  slope  of  the  J_  from  the  slope  of 
the  given  line  by  (a)  and  of  the  parallel  by  (b\ 

Then  the  use  of  the  formula  for  a  line  through  a  given 
point  with  a  given  slope  will  give  the  required  equation. 

Example:     Find  the  equation  of  aj_  to  3^  +  2^=5 


42 


Analytical  Geometry. 


through  the  point  (-  i,  3).     The  slope  of3#  + 
is  —  |  \y  =  —  |  x  +  f  ],  hence  the  slope  of  the  J_  is 


The  type  equation  for  a  line  with  a  given  slope  through 
a  given  point  is        y  —  y'  =  m  (x  —  x'} (C) 

Here  m  =  f ,  xf  =  —  i  and  /  =  3. 

Substituting;         y  —  3  =  $  (x  +  i) 
or  3  y  —  2  x  =  n.* 

ART.  30.     In  Art.  n  it  was  shown  how  the  area  of  a 
triangle  may  be  found  when  the  co-ordinates  of  its  vertices 


Fig.    20. 

are  known.     By  the  equation  for  a  line  through  two  given 
points,  the  equations  of  the  sides  may  now  be  found,  and 

*  Comparing  this  equation  to  the  JL  with  the  original  equation 
it  will  be  seen  that  the  coefficients  of  x  and  y  have  simply  inter- 
changed, and  one  of  them  has  changed  sign,  which  suggests 
a  method  of  writing  the  _L  to  a  line.  See  example  at  end  of 
chapter. 


Analytical  Geometry.  43 

from  them  the  angles  by  formula  (F).  Also  we  may 
erect  J_'s  to  the  sides,  at  any  point.  It  will  now  be  shown 
in  Art.  31  how  the  lengths  of  the  sides  may  be  easily 
obtained. 

ART.  31.  To  find  the  length  of  a  line  between  two  given 
points. 

Let  the  points  be  (#',  yf)  and  (V',  y'),  respectively  A  and 
B  in  Fig.  20. 

Draw  AF  and  BCJ_  to  the  #-axis.  They  are  /  and  /' 
respectively.  OF  =  x'  and  OC  =  —  oc".  Draw  also  AH  || 
to  the  #-axis. 

Then  in  the  right  triangle,  ABH,  AB2  =  AH2  +  BIT2- 
Call  AB,  L  (length  of  AB).  Then  L2  =  (OF  +  OC)2  + 
(BC  -  AF)2  =  (otf  -  x")2  +  (f  -  yy  or  since  (xf-x"Y 


L  =  (x"  -  x7)2  +  (f  -  /)2  (written  symmetrically). 
Example  :  Find  the  distance  between  (i,  —  f  )  and  (f  ,  J). 
Call  the  first  (x',  y'}  and  the  second  (x",  y"). 


Then        L  =  V(|  -  i)2  +  (Jf  f)2  =  VTV  + 


ART.  32.  To  find  the  co-ordinates  of  a  point  which 
divides  a  line  between  two  given  points  into  segments  having 
a  given  ratio. 

Say  the  ratio  is  p  :  q,  the  points  are  (x',  /)  and  (#*,  /') 
(A  and  B  in  Fig.  21)  and  the  required  point  P  (x,  y). 
Draw  BH,  PG  and  AF  _L  to  the  #-axis,  and  AK  ||  to  the 


Then  AF  =  /,  PG  =  y,  and  BH  =  /'.  Also  OF  -  x*, 
OG  =  x,  and  OH  =  x".  Also  AP  :  PB  :  :  p  :  q. 

To  find  PG  and  OG  in  terms  of  (xf,  /)  and  (x",  y") 
PG  =  PN  +  NG  =  PN  +  AF.  (i) 


44 


Analytical  Geometry. 


Since  the  triangles  APN  and  ABK  are  similar,  PN  :  BK 
AP  :  AB, 

that  is,  PN  :  (BH  -  AF)  :  :  AP  :  AB, 

or  PN  :/'-/::/>:  p  +  q. 


or 


pG  =  y  = 


P  +  q 

_  pv"  +  qV 


+  y>  [from 


P  +  q 


Likewise, 
=  OH 


H 


Fig.  21. 


Analytical  Geometry.  45 

If  the  point  is  to  bisect  the  line  then  p  =  q,   and  the 
formulae  become 


and         ,= 


2  p  2 

ART.  33.  To  find  the  distance  from  a  given  point  to  a 
given  line. 

Since  parallel  lines  are  everywhere  equally  distant,  the 
expedient  suggests  itself  of  drawing  a  line  through  the 
given  point  parallel  to  the  given  line,  and  determining  the 
distance  between  these  two  lines  at  the  most  convenient 
point. 

Again,  since  perpendicular  distance  of  course  is  meant, 
the  normal  equation  is  naturally  suggested,  because  it  is 
determined  by  a  perpendicular  from  the  origin. 

Clearly,  since  these  two  lines  are  parallel,  the  angle  /?  in 
the  equation  will  be  the  same  for  both,  and  they  will  differ 
only  in  the  value  of  p.  Also  the  difference  in  the  values  of 
p  for  the  two  will  be  their  distance  apart,  that  is,  will  be 
the  distance  from  the  given  point  to  the  given  line. 

Then  let  x  cos  /?  +  y  sin  ft  =  p,  (E),  be  the  equation  to 
the  given  line  and  x  cos  /?  +  y  sin  ft  =  p'  be  the  equation 
of  a  parallel  line. 

If  this  line  passes  through  the  given  point  (x',  y')  then 
it  must  be  satisfied  by  (x',  y'). 

.'.  xf  cos/?  +  /sin/?=  p'     .....    (2) 
where 

P'-P=  ±d     .......    (3) 

[d  being  the  required  distance].  The  +  sign  will  result 
when  the  point-line  is  farther  from  the  origin  than  the 
given  line;  the  minus  sign,  otherwise. 


46  Analytical  Geometry. 

From  (3),  /=  p  ±d. 

.'.  (2)  becomes  ocf  cos  ft  +  /  sin  /?  =  />  ±  d. 
or  ±  </  =  x'  cos  /?  +  /  sin  /?  -  p (G) 

Since  any  equation  to  a  straight  line  may  be  put  in  nor- 
mal form,  the  above  expression  is  always  applicable.  By 
taking  advantage  of  the  general  form  of  normal  equation, 


"FB5      \/A2  +  B2      x/A2  +  & 
the    formula  (G)  becomes  easier  of  application.     For  in 
above  equations  we  know  that 
^ 

corresponds  to  cos  /?. 


/.   ±d  = 

\2  +B2        x/A2  + 

A^+  5/-C 


This  formula  (G')  may  be  stated  thus: 

To  find  the  distance  from  a  given  point  to  a  given  line, 
put  the  equation  of  the  line  into  the  form  Ax  +  By  —  C  =  o. 

Substitute  for  x  and  y  the  co-ordinates  of  the  given  point 
and  divide  the  left  hand  member  of  the  equation  by  the  square 
root  of  the  sum  of  the  squares  of  the  coefficients  oj  x  and  y. 
The  quotient  is  the  required  distance. 

Example:  Find  distance  from  (—  2,  3)  to  3  x  +  4  y  =  —  9. 

Comparing  Ax  +  ~By  =  C, 
A=  3,  B  =  4,  C=  -  9, 


Analytical  Geometry.  47 

-  C       3  (-  2)  +  4  (3)  ~  (~9) 


VA2  +  B2 
-  6  +  12  +  Q 


Since  it  is  merely  distance  wanted,  the  sign  of  d  is  not 
important. 

SYSTEMS    OF    LINES. 

ART.  34.  Since  parallel  lines  have  the  same  slope,  but 
different  intercepts,  and  since  the  slope  is  determined 
entirely  by  the  coefficients  of  x  and  y,  the  equations  of 
parallel  lines  can  differ  only  in  the  absolute  term. 

Thus  Ax  +  By  =  K  is  the  equation  of  a  line  ||  to  Ax 
+  By  =  C.  Then  two  equations  that  differ  only  in  their 
absolute  terms  represent  parallel  lines. 

Again;  since  the  relation  between  the  slopes  of  perpen- 
dicular lines  is  given  by  the  equation  mf  —  -  — ,  and  m 

and  m'  are  determined  by  dividing  the  coefficient  of  x  by 
the  coefficient  of  y  in  the  equations  of  the  perpendicular 
lines,  if  the  coefficients  of  x  and  y  be  interchanged  and  the 

sign  of  one  of  them  reversed,  the  relation  m'  = will 

m 

be  satisfied.  The  absolute  term  of  course  will  be  different 
in  the  two  equations. 

Thus,  Ex  —  Ay  =  L  is  the  equation  of  a  line  perpen- 
dicular to  Ax  +  By  =  C. 

Again;  (Ax  +  By  -  C)  +  K  (A'*  +  B'y  -  C')  -  o  (i) 
is  the  equation  of  a  line  through  the  intersection  point  of 
A*  +  By  =  C  (2)  and  A'x  +  B'y  =  C'  ...  .  (3) 

For,  transposing  C  and  C'  in  (2)  and  (3), 


48  Analytical  Geometry. 

Ax  +  By  -  C  =  o. 
A'x  +  E'y  -  C'  =  o. 

Let  (xf,  /)  represent  their  intersection  point.  Since 
this  point  is  on  both  lines,  it  satisfies  both  equations;  hence, 

Ax?   +  B/  -  C  =  o (4) 

and  AV  +  B'/  -  C'  =  o (5) 

multiply  (5)  by  K  and  add  to  (4); 

(A*'  +  B/  -  C)  +  K  (AV  +  By  -  CO  =  o         (6) 
If  (V,  yf)  be  substituted  in  (i)  we  get  (6),  but  we  know 
(6)  is  true. 

.'.  (X,  yf)  satisfies  (i),  and  hence  (i)  is  the  equation  of 
a  line  through  (xf,  y).  Since  K  is  an  undetermined  con- 
stant, we  can  get  the  equations  of  any  number  of  lines 
through  (X,  y)  by  giving  K  different  arbitrary  values. 

Example:  To  find  equation  of  a  line  through  the  inter- 
section of  3^—  5  y  =  6  and  2  x  +  y  —  9- 
By  above  formula  the  equation  is, 

(3  x  -  5  y  -  6)  +  K  (2  x  +  y  -  9)  =  o. 
If  the  line  must  also  pass  through  another  point,  say 
(3,  —  i),  K  may  be  determined.     For  substituting  (3,  —  i) 
for  x  and  y, 

(9  +  5  -  6)  +  K  (6  -  i  -  9)  =  o, 
whence        K  =  2  and  above  equation  becomes 

(3  x  -  5  y  -  6)  +  2   (2  x  +  y  -  9)  =  o, 
or  7  x  —  3  y  =  24. 

Example :  Find  the  line  J-  to  #— 3;y=5  through 
(2,  —  i).  Its  equation  by  Art.  34  is 

$x+  y=  k. 

Since  (2,  —  i)  must  satisfy  it,  6  —  i  =  k,  or  k  =  5. 
Hence  3  x  +  y  =  5  is  the  required  line. 


Analytical  Geometry.  49 

EXERCISE   VII. 

1.  Find  the  equation  of  a  line  whose  intercepts  are  —  3 
and  —  5. 

2.  Put  the  following  into  symmetrical  form  and  deter- 
mine their  intercepts. 


-3, 


3.  The  points   (5,   i),   (—2,  3)  and  (i,   —  4)  are   the 
vertices  of  a  triangle.     Find  the  equations  of  its  medians. 

4.  In  Ex.  3,  find  the  equations  of  the  altitude  lines. 

5.  What  are  the  angles  of  the  triangle  in  Ex.  3  ? 

6.  What  is  the  equation  of  the  line  J_  to  2^—3^=5 
through  (—  i,  2)? 

7.  What   is   the   equation   of   line   ||    to   2^—3^=5 
through  (—  i,  2)? 

8.  What     is     the     angle     between    y  +  2  x  =  5     and 
3  y  —  x  =  2? 

9.  The  points  (8,  4)  and  (6,  8)  are  on  a  circle  whose 
centre  is  (i,  3).     What  is  the  equation  of  the  diameter  J_ 
to  the  chord  joining  the  two  points  ? 

10.  What  are  the  co-ordinates  of  the  point  dividing  the 
line  joining  (—  3,  —  5)  and  (6,  9)  in  the  ratio  1:3? 

11.  Prove  that  the  diagonals  of  a  parallelogram  bisect 
each  other. 

12.  Show  that    lines   joining    (3,   o),    (6,   4),    (-  i,   3) 
form  a  right  triangle. 

13.  The  vertices  of  a  triangle  are  (4,  3),  (2,  —  2),  (—3,5). 
Show  that  the  line  joining  the  mid-points  of  any  two  sides 
is  parallel  to,  and  equal  to  J  of,  the  third  side. 


50  Analytical  Geometry. 

14.  Show  that  (-  2,  3),  (4,  i),  (5,  3),  and  (-1,  5)  are 
the  vertices  of  a  parallelogram. 

15.  Show  that  the  line  joining  (3,  —  2)  with  (5,  i)  is 
perpendicular  to  the  line  joining  (10,  o)  and  (13,  —  2). 

16.  (2,  i),   (—4,  —3),  and  (5  —  i)  are  the  mid-points 
of  the  sides  of  a  triangle.     What  are  its  vertices? 

17.  Three  of  the  vertices  of  a  parallelogram  are  (2,  3), 
(-4,  i),  (-  5,  -  2).     What  is  the  fourth? 

18.  Find  the  point  of  intersection  of  the  medians  of  the 
triangle  whose  vertices  are  (i,  2),  (—  5,  —3),  (7  —  6). 

19.  What  is  the  distance  from  the  point  (—  2,  3)  to  the 
line  5  #  =  12  y  —  7? 

20.  Find  the  distance  between  the  sides  of  the  parallelo- 
gram in  Ex.  14. 

21.  Change  3  #  —  4  ;y  =  5  to  the  normal  form. 

22.  Find  the  co-ordinates  of   the   points   trisecting   the 
line  joining  (2,  i)  and  (—  3,  —  2). 

23.  Find  the  distance  from  (2,  5)  to  2  x  —  3  y  =  6. 

24.  Find  the  altitude  and  base  of  the  triangle  whose  vertex 
is  (3,  i)  and  whose  base  is  the  line  joining  (|,  i)  and  (4,  — f ). 

25.  Find  the  area  of  the  quadrilateral  whose  vertices  are 
(6,8),   (-4,0),   (-2, -6),   (4, -4). 

26.  Find  the  angles  of  the  parallelogram  whose  vertices 
are  (i,  2),   (-  5,  -3),   (7,  -  6)   (i,  -  n). 

27.  One  side  of  an  equilateral  triangle  joins  the  points 
(2,V3)  and  (~  Ij  4  v  3)-     What  are  the  equations  of  the 
other  sides? 

28.  What  is  the  equation  of  a  line  passing  through  the 
intersection  of  the  lines  3  x  —  y  =  5,  and  2  x  +  3  y  =  7 
and  the  point  (—  3,  5)? 

29.  By  Art.  34,  find  the  equations  to  the  medians  of  the 
triangle    whose    sides    are    y=2x-\-ity-\-x-\-i=Q 
and  5  x  =  2  y  -f  2. 


Analytical  Geometry .  51 

30.  Find  the  co-ordinates  of  the  centre  of  the  circle  cir- 
cumscribing the  triangle  whose  vertices  are  (3,  4),  (i,  —2), 

(-1,2). 

31.  The  base  of  a  triangle  is  2  b  and  the  difference  of  the 
squares  of  the  other  two  sides  is  d?.     Find  the  locus  of  the 
vertex. 


CHAPTER   IV. 

TRANSFORMATION    OF    CO-ORDINATES. 

ART.  35.  It  sometimes  simplifies  an  equation  to  change 
the  position  of  the  axes  of  reference  or  even  to  change  the 
inclination  of  these  axes  from  a  right  to  an  oblique  angle, 


-X' 


Fig.  22. 

or  both.  To  accomplish  this  it  is  only  necessary  to  express 
the  original  co-ordinates  of  any  point  on  the  line  in  terms 
of  new  co-ordinates  determined  by  the  new  axes  and  neces- 
sary constants. 

ART.  36.  To  change  the  position  of  the  origin  without 
changing  the  direction  of  the  axes  or  their  inclination. 

Let  P  be  any  point  on  a  given  line  whose  equation  is  to 
be  transformed. 

Let  its  co-ordinates  be  x  =  OC  and  y=  PC  (Fig.  22), 
52 


Analytical  Geometry.  53 

referred  to  the  axes  OX  and  OY.  Let  O'X'  and  O'Y'  be 
new  axes,  such  that  the  origin  O'  is  at  the  distance  O'A  =  a, 
from  the  axis  OY,  and  at  the  distance  O'B  =  b,  from  OY. 

Extend  PC  to  D  _\_  to  O'X',  since  the  direction  of  the 
axes  is  not  changed. 

Then  the  co-ordinates  of  P  with  respect  to  the  new  axes 
are  xf  =  O'D  and  /  =  PD. 

Now,  OC  =  AD  =  O'D  -  O'A,  or  x  =  x'  -a 


PC  =  PD  -  CD  =  PD  -  O'B,  or  y  =  yf  -  b     l 

It  will  be  observed  that  (— 0,  —  b)  are  the  co-ordinates  of 
the  new  origin  referred  to  the  old  axes,  hence  the  old  co-or- 
dinates are  equal  to  the  new  plus  the  co-ordinates  of  the 
new  origin,  plus  being  taken  in  the  algebraic  sense. 

Example:  What  will  the  equations2—  43;  + y2  —  67=3 
become,  if  the  origin  is  moved  to  the  point  (2,  3),  direction 
being  unchanged  ? 

Here,  x  =  xf  +  2  and  y  =  y'  +  3- 

Substituting, 

(yf  +  2)2  -  4  (*'  +  2)  +  6"+3)2-6  (/+3)  =  3. 
Expanding    and    collecting,    x'2  -{-y  /2  =  16    or    dropping 
accents;  x2  +  y2  =  16,   which  indicates  how  an  equation 
may  be  simplified  by  transferring  the  axes. 

ART.  37.  To  change  the  direction  of  the  axes,  the  angle 
remaining  a  right  angle. 

Let  O'X"  and  O'Y"  be  the  new  axes,  the  axis  O'X"  mak- 
ing the  angle  0  with  the  old  X-axis,  and  the  new  origin  O' 
being  at  the  point  (a,  b). 

Let  the  old  co-ordinates  of  P  [OD  and  PD  in  the  figure] 
be  (x,  y)  and  the  new  co-ordinates  [O'A  and  PA  in  the 
figure]  be  (V,  /).  Draw  O'C  and  BA  ||  to  OX  and  AE  J_ 
to  OX,  then  Zs  AO'C  and  BPA  both  equal  0. 

OD  =  x  =  OF  +  O'C  -  BA      .     .     .     (i) 


54 


Analytical  Geometry. 

In  the  right  triangle,  AO'C,  O'C,  =  O'A  cos  AO'C 
[by  Trig.].     That  is,  O'C  =  yf  cos  6. 


Fig-  23- 

Also  in  BPA,  BA  =  PA  sin  BPA  or  BA  =  /  sin  0;  and 
OF=  a. 

Substituting  in  (i), 

x  =  a  +  #'  cos  0  —  y  sin  0. 

Again:         PD  =  y  =  O'F  +  AC  +  PB   .     .     .     .     (2) 
O'F  =  b;     AC  =  O'A  sin  AO'C  or  AC  =  yf  sin  0 

and  PB  =  PA  cos  BPA  or  PB  =  /  cos  0. 

Substituting  in  (2), 

y  =  b  4-  xf  sin  0  -f  yf  cos   0  )  (^-^ 

x  =  a  +  xf  cos  0  —  y  sin   0  ^ 

If  in  any  equation  these  values  be  substituted  for  x 
and  y,  the  resulting  equation  will  represent  the  same  locus 
referred  to  axes  inclined  at  the  angle  0  to  the  old  X-axis, 


Analytical  Geometry.  55 

with  the   origin  at   (a,  £).     As  a  rule  the  origin  remains 
the  same,  hence  a  =  o,  b  =  o,  and  (K)  becomes, 

y  =  x'  sin  0  +  y'  cos  0    ) 
x  =  x'  cos  $  —  y'  sin  #    k 

Example:   What   does   equation    3  x  —  2  y  —  5    become 
when  the  inclination  of  the  axes  is  changed  30°? 
Here  sin  30  =  J;  cos  30  =  J  \/3 

and  7=  iaM- JVSY, 


Substituting,  3 
or  (I  VT~  i)  X?  -  (t  +  V3)  /  =  5- 

ART.  38.  A  very  similar  procedure  in  the  case  where 
the  axes  are  changed  from  rectangular  to  oblique,  and  the 
origin  moved  to  the  point  (a,  b),  gives  rise  to  the  formulae, 

y  =  b  +  x'  sin  6  +  yf  sin  (f>  )  ,^^ 

x  =  a  -\-  x'  cos  6  +  y'  cos  <£  J 

0  and  0  being,  respectively,  the  angles  made  by  the  new 
Y-axis  and  Y-axis  with  the  old  X-axis. 
When  the  origin  is  not  changed, 

a  =  o  and  b  =  o,  and  (J)  becomes 
y  =  x'  sin  0  +  y'  sin 
x  =  x'  cos  6  +  y'  cos 

ART.  39.  To  change  the  co-ordinates  from  rectangular 
to  polar. 

The  method  is  entirely  similar  to  the  foregoing;  the  find- 
ing of  the  simplest  equational  relation  between  the  old 
and  the  new  co-ordinates,  using  necessary  constants. 

In  Fig.  24,  let  O'  be  the  pole  and  O'N  the  initial  line, 
the  co-ordinates  of  O'  being  (a,  b);  the  rectangular  co- 
ordinates of  P  being  (x,  y)  and  the  polar,  (r,  #),  respec- 


Analytical  Geometry. 


tively,  OB,  PB,  O'P,  and  Z  PO'N  in  the  figure.  The  angle 
between  the  initial  line  and  the  X-axis  is  <j>. 

It  is  then  simply  a  question  of  expressing  x  and  y  in 
terms  of  r,  6  and  (f>. 

The  right  triangle  usually  supplies  the  simplest  relations, 
so  we  draw  O'AJJo  PB,  giving  us  the  right  triangle  PO'A 
involving  r,  6  and  O'A  =  FB,  a  part  of  x. 


Fig.  24- 

Then  OB  =  x  =  OF  +  FB  =  OF  +  O'A, 

or  x  =  a  +  r  cos  (0  +  <£>) 

[since  O'A  =  O'P  cos  PO'A  =  r  cos  (0  +  0)]. 
Also,         PB  =  y  =  AB  +  PA  =  O'F  +  PA, 

y  =  b  +  r  sin    (0  +  </>)  ^ 
^-=  a  -f  r  cos   ((9  +  0)  $ 

If  the  initial  line  is  ||  to  the  X-axis,  0  =  o  and  (M)  becomes 
^  =  b  +  r  sin  0  )  _  ^M/^ 

jc:  =  a  4-  r  cos  0  ( 


or 


(M) 


A  nalytical  Geometry.  5  7 

If  the  pole  is  at  the  origin,  a  =  o  and  b  =  o 

y=r  sin  °  }  CM."\ 

and  X=rcosO    \ 

ART.  40.     To  change  from  polar  to  rectangular  co-ordi- 
nates. 

It  is  here  necessary  only  to  solve  equations  (M"),  say, 
for  r  and  0,  as  (M")  gives  the  usual  form. 
Thus,  squaring  equations  (M"), 

y2  =  r2  sin2  6 
x*  =  r2  cos2  6. 

Add;        x2  +  y2  =  r2  (sin2  6  +  cos2  6)  =  r2 
[since  sin2  0  +  cos2  0  =  i]. 

Dividing  the  first  equation  in  (M")  by  the  second, 


#        r  cos  o 
Example:  Change  to  rectangular  form 


Substituting  in  above  equation,  remembering  that 
cos  2  0  =  cos2  6  -  sin2  6  =  cos2  6  (i  -  tan2  6) 

i  —  tan2  6  _     i  —  tan2  6 
sec2  6  i  +  tan2  0  : 


or, 


58  Analytical  Geometry. 

EXERCISE    VIII. 
Transformation  of  Co-ordinates. 

1.  What    does   y2  =  2  px   become    when    the    origin    is 

moved  to  (  —  *-  >    o    J  without  changing  the  direction  of  the 
\      2          / 

axes? 

2.  What    does    a2y2  +  b2x2  =  a2b2    become    when     the 

origin  is  moved  to    (  -  — ,  o  J ,  axes  remaining  parallel  ? 

3.  What  does  y2  +  x2  +  4  y  —  4:^  —  8  =  o  become  when 
origin  is  moved  to  (2,  —  2)  ? 

4.  What  does  y2  =  Sx  become  when  the  axes  are  turned 
through  60°,  origin  remaining  the  same? 

5.  What   does   y2  =  2  px   become   when    the   origin   is 
moved  to  the  point  (/»,»)? 

6.  What    does    a2y2  +  b2x2  =  a2b2    become    when    the 
origin  is  moved  to  (h,k)? 

7.  What  does  2  \/3  x  +  2  y  =  9  become  when  the  axes 
are  turned  30°,  origin  remaining  the  same  ? 

8.  What    does     b2x2  —  a2y2  =  a2b2    become    when    the 

Y-axis  is  turned  to  the  right,  cot  -1  — »    and  the  X-axis  to 

a 

the    right,    tan  -1  [observe  negative  angle]  ? 

a 

9.  Transform  the  polar  equation  p  =  a   (1+2  cos   6) 
to  a  rectangular  equation  with  the  origin  at  the  pole,  and 
the  initial  line  coincident  with  the  X-axis. 

10.  Change  (x2  +  y2)2  =  a2(x2  —  y2)  to  the  polar  equa- 
tion under  the  conditions  of  Ex.  9. 

11.  Change  p2  =    — to    rectangular  co-ordinates, 

COS  2  6 

conditions  remaining  the  same. 


Analytical  Geometry.  59 

12.  Change    to    rectangular    co-ordinates,    under   same 

conditions,  p  =  a  sec2  —  • 

2 

13.  p  =  a  sin2#. 

14.  p  =         —£ — -  . 

i  —  cos  a 

15.  Change  to  polar  co-ordinates,  under  same  conditions. 


16.  4  a2je  =  2  ay*  —  ^2. 

17.  #3  +  3^  =  a*. 

18.  4  jc2  +  9  /  =  36. 


CHAPTER  V. 
THE    CIRCLE. 

ART.  41.     To  find  the  equation  to  the  circle. 

Remembering  the  definition  for  the  equation  of  a  locus, 
namely,  that  it  must  represent  every  point  on  that  locus,  it 
is  only  necessary  as  usual  to  find  the  relation  between  the 
co-ordinates  of  any  point  on  the  circle  in  terms  of  the  ne- 
cessary constants,  which  are  plainly  in  this  case,  the  co-ordi- 
nates of  the  centre  and  the  radius. 

Let  P  be  any  point  on  the  circle  A,  the  co-ordinates  of 
whose  centre  are  (h,  k).  The  condition  determining  the 


Fig.  25. 

curve  is  that  every  point  on  it  is  equally  distant  from  its 
centre.  Draw  the  co-ordinates  of  P  [PC,  OC]  and  call 
them  (x,  y),  also  AB  J_  to  PC,  forming  the  right  triangle 
APB,  involving  r  and  parts  of  x  and  y. 

60 


Analytical  Geometry.  61 

Then  AB2  +  PB2  =  AP2     .....     .     .     (i) 

AB  =  DC  =  OC  -  OD  =  x  -  h, 
PB  =  PC  -  BC  ==  PC  -  AD  =  y  -  k. 
Substituting  in  (i):  (x  -  h)2  +  (y  -  k)2  =  r2  .     .      (L) 
Performing  indicated  operations  in  (L)  and  collecting, 

x2  +y2  -  2hx-  2ky=  r2  -  h2  -  k2. 
Calling  -  2  h,    m\  -  2  k,    n    and     (h2  +  k2  -  r2),    R2 
for  simplicity,  (L)  becomes, 

x2  +  y2  +  mx  +  ny  +  R2  =  o  .    .    .    .    (L') 

It  is  evident  from  (L')  that  any  equation  of  the  second 
degree  between  two  variables  in  which  no  term  containing 
the  product  of  the  variable  occurs,  and  where  the  coefficients 
of  the  second  power  terms  are  either  unity  or  both  the 
same,  is  the  equation  of  a  circle. 

Putting  (L')  in   the  characteristic  form   (L)  by  adding 

m2        n2 

to  both  sides  —  -  +  —  , 
4  4 

we  have,        x2  +  mx  +  --  f-  y2  +  nx  +   — 

4  4 

-  ^!  _L  »!  _  R2 

*•  ) 

4          4 
or,  (*+-)2+  (y+  ^-)2 

2  2 

m 


2 


,  p2 

—  —  -)  --  —  K   — 


44  4 

Comparing  with  (L),  we  find 

i,  -     ~  m     i,  -         n  .    2         w2+  ^2  —  4  R2 
"  ---  ,  K  —  --  -,  r  =    -  —  -  • 

2  4 

That  is,  the  co-ordinates  of  the  centre  are  (—  —  ,  —  —  ), 

2  2 

and  the  radius  is       i  \/m2  +  n2  -  4  R2. 


62  Analytical  Geometry. 

Example:  Find  the  co-ordinates  of  the  centre  and  the 
radius  of  x2  +  y2  —  2  x  +  6  y  —  26  =  o. 

Comparing  this  with  (L/),  x2  +  y2  +  mx  +  ny  +R2  =o, 
we  find,  m=  —  2,  n  =  6,  R2  =  —  26;  hence  the  co-ordi- 
nates of  the  centre, 


/        m              Wx              ,  -   2                   v           /               \ 

(  --  ,  -    -),are(-  -,  --  )  =  (i,  -3), 

22  22 

and  the  radius 


-  4  R2 


=  i  V4  +  36  -  (- 104) 
=  i  \7i44  =  6. 

This  equation  put  in  form  (L)  would  be, 

(x  -  i)2    +  (y  +  3)2  -  36. 

ART.  42.  As  it  takes  three  conditions  to  determine  a 
circle,  and  as  the  above  equations  contain  three  arbitrary 
constants,  if  three  conditions  are  given  that  will  furnish 
three  simultaneous  independent  •  equations  between  these 
constants,  their  values  can  be  found,  and  hence  the  equation 
to  the  circle. 

The  three  conditions  may  be,  for  instance,  three  given 
points  on  the  circle,  or  two  given  points  and  the  radius,  etc. 

Example:  Find  the  equation  for  the  circle  passing  through 
the  points  (3,  3),  (i,  7),  (2,  6). 

Taking  the  general  equation, 

x2  +  f  +  mx  +  ny  +  R2  =  o  .  .  .  (I/) 
these  three  points  must  each  satisfy  this  equation  if  it  is  to 
represent  the  circle  passing  through  them,  since  they  are 
on  it.  Hence,  substituting  them  successively  for  x  and  y 
in  (L'),  we  get  three  equations  between  m,  n  and  R2  as 
follows: 


Analytical  Geometry.  63 

9  +  9    +3w  +  3W  +  R2 

i  +  49  +     m  +  7  n  +  R2  =  o     or 

4  +  36  +  2  w  +  6  w  +  R2  =  o 

3/w  +  $n  +  R2=    -  18 (i) 

m  +  7  w  +  R2  =  -  50 (2) 

2m  +  6n  +  R2=  -  40 .    .    (3) 

Subtract  (2)  from  (i)  and  (2)  from  (3). 

2  m  —  4  «  =  32  or  m  —  2  n  =  16     .    .    .    (4) 
m  -      n  =  10     ...    (5) 
Subtract  (5)  from  (4);  n  =  —  6. 

whence  w  =  4, 

and  R2  =  —  12. 

Substituting  these  values  of  the  constants  in  (L'), 

x*  +  ;y2  +  4  x  —  6  y  —  12=0, 
the  required  equation. 

ART.  43.     When  the  origin  is  at  the  centre  of  the  circle, 
h  and  k  are  both  zero,  and  the  equation  becomes, 

x>+f=r*      (L") 

which  is  the  form  usually  encountered. 

ART.  44.     The  polar  equation  is  readily  derived  from 
(L)  by  making  the  substitutions  for  transformation  from 
rectangular    to    polar  co-ordinates,  taking    the    X-axis    as 
initial  line  and  the  pole  at  the  origin. 
Then  y  =  p  sin  0, 

x  =  p  cos  6, 
k  =  Pf  sin  0', 
h=  p'  cos  #', 

where  (p,  6}  are  the  polar  co-ordinates  of  any  point  on  the 
circle    and  (pf,  0'}  are  the  polar  co-ordinates  of  the  centre. 
Making  these  substitutions  in  (L),  we  get  : 

(p  cos  0  -  Pf  cos  0')*    +  (p  sin  0  -  p'  sin  O*)2  =  r\ 
or,  p2  cos2  0  -  2  pp'  cos  0  cos  0'  +  p/2  cos2  d'  + 
p2  sin2  0  —  2  pp'  sin  0  sin  d'  +  />2  sin2  0'  =  r2. 


64  Analytical  Geometry. 

Collecting,  /o2(cos2  0  +  sin2  0)  +  p'2  (cos2  0'  +  sin2  6') 
-  2  pp'  (cos  0  cos  6'  +  sin  0  sin  0')  =  r2. 
whence 

p2  +  p'2  -  2  ppf  cos  (0  -  00  -  r2 
[since          cos2  0  +  sin2  0  =  i 
and  cos  6'  cos  0'  +  sin  0  sin  0r=  cos  (0  -  0')]. 

TANGENTS    AND    NORMALS. 


ART.  45.  To  find  the  equation  of  a  tangent  to  the 
circle  x2  +  y2  =  r2.  Since  a  line  may  be  determined  by 
two  conditions,  and  a  tangent  must  be  perpendicular  to  a 
radius  and  touch  the  circle  at  one  point,  the  radius  being 
in  this  case  the  distance  from  the  origin  to  the  line  furnishes 
one  condition  and  the  point  of  tangency  another. 

Knowing  the  equation  to  a  line  determined  by  two  points, 

(*"*•') 


Fig.  26. 

and  taking  these  two  points  on  the  circle,  we  are  able  to 
convert  this  condition  in  the  special  case  of  the  tangent 
into  the  point  of  tangency  and  the  distance  from  the  origin. 
The  equation  of  a  line  through  two  points  (xf,  y'}  and 
(*",/)  is, 


Analytical  Geometry.  65 

Let  these  two  points  be  B  and  C  on  circle  O,  then 
(x',  /)  and  (V',  /')  must  satisfy  the  equation  to  the 
circle;  hence 

x'z  -f2   =  rz      ......      2 


If  these  conditions  be  imposed  on  (V,  y')  and  (V',  y)  in 
equation  (B),  it  will  become  a  secant  line  to  the  circle. 

Subtracting  (2)  from  (3), 

x"2  -  x'2  +  y"2  -  y'2  =  o, 
or,  x"2-x'2  =  -  (y"2-/2)', 

factoring,         (**  -  *')   (x»  +  xf)=-  (/-/)  (/'+/), 

y/  _  y          y*  +  X' 
whence  '—  -  •-=  --  —  —  -. 

xf  —  x'  y"  +  yr 

Comparing    (B)   with   the   equation   to   a   straight   line 
having  a  given  slope  and  passing  through  a  given  point, 


.     .     .     (B) 
y  -  yf  =  m  (x  -  x'} (C) 

It  is  evident   that   •- 2L   =  m    so  that  the  slope  of  a 

x"  —  x' 

line  through  two  given  points  (xf,  y'}  and  (x",  y")  is  repre- 

y"  -  y' 
sented  by       „_ — ;  • 

Hence  the  value  of    ?    ~  ?       —    x     *" x       represents 

x"  -  xf  y"  +  y' 

the  slope  of  a  secant  line  to  the  circle,  and  if  this  value 
be  substituted  in  (B)  the  result  will  be  the  equation  of  a 
secant  line  through  the  point  (xf,  y')  with  the  slope 


66  Analytical  Geometry. 

Then  if  (#",  /')  is  taken  nearer  and  nearer  to  (V,  /) 
the  secant  will  approach  the  position  of  the  tangent  at 
(^  /),  and  when  (V',  /')  coincides  with  (V,  /)  it  will 
be  the  tangent.  Clearly  we  are  at  liberty  to  take  (of,  /') 
where  we  please,  since  it  was  any  point  on  the  circle. 

Substituting  in    (B),  y  -  /  =  + 


Making     x?  =  x'  and  f  =  y', 

y  -  y'  =  -  ^  (x  -  *0  =  -  y  (x  -  *'); 

clearing  of  fractions,  yy'  —  y'2  =  —  xx'  +  xn] 

transposing,  xxf  +  yy'  =  x'2  +  /2. 

But  by  (2),  x'2  +  y'2  =  r2. 

...  xtf  +yy'=r2        ...  .      (Tc) 

Evidently  it  would  serve  as  well  to  make  (V,  y')  approach 
(x",  y"},  only  the  line  would  then  be  tangent  at  (x",  /'). 
In  (Tc)  the  accented  variables  always  represent  the  point 
of  tangency. 

Example:  What  is  the  equation  of  the  tangent  to  the 
circle  x2  +  y2  =  10  at  (—  i,  3)? 

Here  r2  =  10,  yf  =     -  i  and  /  =  3. 

Substituting  in  (Tc),  —x  +  $y  =  10  or  3  y  -x  -  10  =o. 

Observe  that  (V,  /)  is  point  of  tangency,  not  (x,  y)\ 
never  substitute  the  co-ordinates  of  point  of  tangency  for 
the  general  co-ordinates  x  and  y. 

Again:  find  equation  of  tangent  to  the  circle  x2  +  y2  =  9, 
from  the  point  (5,  7!)  outside  the  circle. 

The  equational  form  is,  xxf  +  yy'  =  9  .  .  .  .  (i)  and 
it  remains  to  find  point  of  tangency  (V,  /).  The  point 
(S>  7i)  Deing  on  tnis  tangent  must  satisfy  its  equation,  but  it 
is  not  the  point  of  tangency  and  must  not  be  substituted  for 


Analytical  Geometry.  67 

(x',  /).  Hence  substituting  in  (i),  —  5  a/  +  V5  /  =  9.  (2) 
Also,  since  (:*;',  /)  is  on  the  circle  it  must  satisfy  circle 
equation;  that  is, 

*"+/2=_9 (3) 

Combining  the  simultaneous  equations  (2)  and  (3),  we  get, 

That  is,  there  are  two  tangents,  as  we  know  by  Geometry; 
namely,  63  x  —  16  y  =  195  and  4^  —  3^=  15.  [Gotten 
by  substituting  these  values  of  (xf,  y')  in  (Tc).] 

CIRCLE. 

ART.  46.  To  express  the  equation  of  a  tangent  to  a 
circle  in  terms  of  its  slope. 

Evidently  the  tangent  being  a  simple  straight  line  may 
be  determined  by  its  slope  as  well  as  by  the  point  of  tan- 
gency,  if  the  slope  be  such  that  the  line  will  touch  the  circle. 

Hence  it  is  a  question  of  determining  this  n'ecessary  value 
of  m.  If  we  take  the  general  slope  equation  to  a  straight 
line  and  find  a  relation  between  m,  b  and  r  such  that  the 
line  will  touch  the  circle  of  radius,  r,  it  is  sufficient. 

Again,  regarding  the  tangent  as  the  limiting  position  of 
the  secant  line,  as  its  two  points  of  intersection  with  the 
circle  approach  coincidence  (as  in  Art.  45),  if  we  combine 
the  slope  equation  of  a  straight  line  with  the  equation  to  a 
circle,  we  get  in  general  their  two  points  of  intersection 
expressed  in  the  constants  they  contain;  if  then  we  deter- 
mine (by  Algebra)  the  conditions  these  constants  must 
fulfil  among  themselves  that  the  two  points  of  intersection 
shall  coincide,  or  become  one  point,  we  have  the  desired 
result. 

Let  y  =  mx  +  b,  (i)  be  the  slope  equation  of  a  straight 
line,  and  x2  +  y2  =  r2,  (2)  be  the  equation  to  a  circle. 


68  Analytical  Geometry. 

Regarding  (i)  and  (2)  as  simultaneous,  and  substituting 
the  value  of  y  from  (i)  in  (2),  we  get  a  quadratic  in  x, 
whose  two  roots  are  the  abscissas  respectively  of  the  two 
points  of  intersection. 

We  get  then,     x2  +  (mx  +  b)2  =  r2, 

x2  +  m2x2  +  2  mbx  +  b2  =  r2, 

(i  +  m2)  x2  +  2  mbx  +  (b2  -  r2)  =o.  (3) 

By  the  theory  of  quadratics  in  algebra  we  know  that  the 
two  values  of  x  will  be  the  same  in  (3 )  if  it  can  be  separated 
into  two  equal  factors,  that  is,  if  it  is  a  perfect  square. 

By  the  binomial  theorem  it  will  be  a  perfect  square 
if  the  middle  term  is  twice  the  product  of  the  square  roots 
of  the  first  and  last  terms  (like  a2  +  2  ab  +  b2). 

Hence  (3)  will  have  two  equal  values  of  x  (that  is,  equal 
roots)  if 


2  mbx  =  2  v/(i  +  m2)  (b2  —  r'2)  x~, 
or  squaring;  if  4  m2  b2x2  =  4  (i  +  m2}  (b2  —  r2)  x2  = 

4  (b2x2  -  r2x2  +  b2m2x2  -  r2m2x2), 
dividing  by  4  x2;     b2m2  =  b2  —  r2  +  b2m2  —  r2m2, 

b2  =  r2  +  r2m2  =  r2  (i  +  m2), 

or  b  = 


If  this  condition  be  fulfilled,  clearly  the  equation  of  the 
secant  y  =  mx  +  b  will  become  the  equation  of  the  tangent 
y  =  mx  ±  r\/i  +  m2      ...     (Tc>  m) 

The±  sign  indicates  that  there  will'  be  two  tangents  with 
the  same  slope,  as  should  be  the  case,  having  ^-intercepts 
numerically  equal,  but  opposite  in  sign,  or  vice  versa. 

Example  :  Find  the  value  oi  b  in  y  =  &  x  +  b,  that  the 
line  may  be  tangent  to  the  circle  x2  +  y2  =  25. 


Analytical  Geometry.  69 

Bv  condition  formula,  b  =  ±  r  ^/i  +  w2, 


we  must  have,    b  =  ±  5  \/i  +   64  =  ± 

225  T   225 

Hence  the  equations  of  the  tangents  are 


i5  3  i5  3 

or      15  y  =  8  x  +  85  and  15^=8  x—  85. 

ART.  47.  The  normal  to  any  curve  at  a  specified  point 
is  defined  as  the  line  perpendicular  to  the  tangent  at  that 
point. 

It  is  evident  from  geometry  that  the  normal  to  the  circle 
at  any  point  is  the  radius  drawn  to  that  point. 

Since  the  normal  is  perpendicular  to  the  tangent,  if  the 
slope  of  the  tangent  is  known  the  slope  of  the  normal  is 

readily  found    lm'= )>   and  as  it  must  pass  through 

the  point  of  tangency,  we  have  all  the  conditions  necessary 
to  determine  its  equation. 

To  find  the  equation  of  the  normal  to  the  circle  x2  +  y2=  r2. 
Let  the  point  of  tangency  be  (V,  /).  The  equation  to 
the  tangent  at  this  point  is  xxr  +  yyf  =  r2,  or  in  slope  form, 

y  =  -  —  x  +  -    (i),  and  its  slope  is  -  —  • 

/  /  / 

Since  the  normal  is  perpendicular  to  it,  its  slope  is 

-—  =  '- 


— x 


x' 


y' 

The  equation  of  a  line  through  (V,  /)  with  slope  m'  is 
y  -  y'  =  mf  (x  -  xf)   .    .    .     .    [by  (C)] 

But,  mf  is  here  equal  to  — , 

'V 


70  Analytical  Geometry. 

hence  the  normal  equation  is      y  —  y'  =  -,  (#  —  #0, 

oc 

or  xfy  —  x'y'  =  xy'  —  x'y', 

whence  y  =  t.f  x     .........     (Nc) 

x 

This  may  be  written  in  slope  form,  using  the  slope  of  the 
tangent,  m,  by  substituting  for  ?!_,  the  slope  of  the  normal, 

its  value   —  —  • 
m 

x 


or  my  +  x  =  o. 

ART.  48.  To  find  the  length  of  a  tangent  from  any  point 
to  the  circle  x2  +  y2  =  r2. 

By  Art.  31,  if  (xlt  y^}  be  the  given  point  and  (V,  y') 
the  point  of  tangency,  the  length  (d)  of  a  line  between  them 
is,  d2  -  K  -  x'}2  +  &-  y')2  =  x2  +  y2  -  2  (x,x'  + 
ytf)  +  x'2  +  y'2,  but  if  (x',  y'}  is  on  the  circle  and  (xlt  y^} 
on  the  tangent,  x'2  +  y'2  =  r2  and  x^xf  +  ytf  =  ^2. 

.-.        ^  =    x*    +  ^2   _    2  f2    _|_   r2  =    ^2    +  y2   _    r2  §         (Dc) 

If  the  origin  is  not  at  the  centre  of  the  circle,  it  is  easy 
to  show  in  exactly  the  same  way  from  equation  (L),  that 


d  =  V(Xl  -  h}2  +  (ft  -  k}2  -  r2  . 

ART.  49.  The  locus  of  points  from  which  equal  tangents 
may  be  drawn  to  two  given  circles  is  called  the  radical 
axis  of  these  circles.  Having  the  above  expression  for 
the  length  of  a  tangent  to  any  circle,  it  is  only  necessary  to 


Analytical  Geometry.  71 

equate  the  two  values  of  d  for  the  two  given  circles,  in  order 
to  find  the  equation  to  the  radical  axis. 
Let  the  circles  be, 
(,_A).  +  (y_i)._fr(Ci)  j      dl 

(,_m)2  +  (y_B)2=R2j(C2)   \  < 

be  any  point  on  the  radical  axis  to  these  circles. 

If  dl  and  d2  are  the  tangent  lengths  from  (xlt  y±)  to  (Q) 
and  (C2)  respectively,  then, 


dL  =  \/(x1  -  h)2  +  (yl  -  k)2  -  r2 
and  d.2  =  v'K  -  O2  +  (ft  -  »)2  ~  R2- 

But  d^  =  d2  or  ^2  =  d2. 

...    (^   _   A)2    +    (^   _   ^)2  _   ^2  =     ^  _   m), 

+  (^-W)2-R2 (3) 

Since  (xlt  y^  substituted  in  the  equation 
(x  -  h)2  +  (y-  k)2  -r2=  (x-  m)2  +  (y-n)2  -  R2  (4) 
gives  (3)  which  we  know  to  be  true,  then  (xlt  yt)  satisfies  (4). 

But  (jCj,  yv)  is  any  point  on  the  radical  axis,  hence  every 
point  on  that  axis  satisfies  (4),  and  /.  (4)  is  the  equation 
of  the  radical  axis  to  (Q)  and  (C2). 

SUBTANGENT   AND    SUBNORMAL. 

ART.  50.  The  Subtangent  for  any  point  on  a  curve  is 
the  distance  along  the  rv-axis  from  the  foot  of  the  ordinate 
of  the  point  of  tangency  to  the  intersection  of  the  tangent 
with  that  axis. 

The  Subnormal  for  any  point  on  a  curve  is  the  distance 
measured  on  the  #-axis  from  the  foot  of  the  ordinate  of 
the  point  of  tangency  to  the  intersection  of  the  normal 
with  that  axis. 

Let  O  [Fig.  27]  be  a  circle,  PT  a  tangent  at  P  (X,  /), 
OP  a  normal  at  the  same  point,  PA  the  ordinate  (/)  of  P. 
Then  AT  =  subtangent  and  OA  =  subnormal  for  P. 


72 


Analytical  Geometry. 


To  find  their  values,  it  is  to  be  observed  that  the  subtangent 
AT  =  OT  -  OA.  OT  =  the  ^-intercept  of  the  tangent, 
which  is  found  as  in  any  other  straight  line  by  setting 


or 
Also, 


Fig.  27- 

y  =  o  in  its  equation  (y  =  o  being  the  ordinate  of  the 
point  T).  Then  in  equation  (Tc)  setting  y  =  o,  we  get 
xx'  +o=r\ 

x  =  OT  =:    r-f . 
OA  =  xf. 

/y*t  yy»'  /\rf 

The  subnormal,  OA  =  xf  evidently. 

Example:   The   subtangent   for   the   point    (3,   4)    on   a 

circle  is  - .     What  is  the  equation  of  the  circle? 

3 

Here  xf  =  3,  /  =  4  and 

From  this  last  equation 


-      — —  =  — , 

3  3 

v/hence  r2  -—  25;        ^  —  5- 

Then  the  equation  to  the  circle  is  x2  +  y1  =  25. 


Analytical  Geometry.  73 

The  origin  is  taken  at  the  centre  of  the  circle  in  these 
discussions  because  that  is  the  usual  form  encountered, 
and  the  processes  are  exactly  the  same  wherever  the  origin 
may  be;  the  greater  simplicity  of  results  recommending 
this  form  of  equation  for  explanation. 

INTERSECTIONS. 

ART.  51.  By  what  has  been  said  in  general  about  the 
intersections  of  lines,  it  follows  that  if  two  circles  intersect, 
the  points  of  intersection  will  be  readily  found  by  combining 
the  two  equations  as  simultaneous.  If  the  circles  are 
tangent,  the  unknowns  x  and  y  will  have  each  one  value, 
or  rather  each  will  have  its  values  coincident. 

Example:  Find  where 

(  x2  +  y2  —  4  #+  2  y  =  o   (i)    )     .  „ 

}    -   ,/.  intersect. 

{x2  +  y2  -  2  y=4  (2)    J 

Subtracting  (i)  from  (2),  4  x  —  4  y  =  4, 
or  x  -  y  =  i       .     .     .     .      (3) 

Substituting  value  of  x  from  (3)  [x  =  y  -f-  i]in  (2), 
y2  +  2y  +  i  +  y2  —  2  y  =  4, 


whence  from  (3),     x  =  i  ±  \/|. 
The  points  of  intersection  are  then  (i  +  \/f  ,  ViT 

(i  -  VI  -  VI). 

Plot  the  figure  and  verify  results. 

(3)    Is   evidently   the   common  chord,    for   both   points 
satisfy  it,  and  it  is  the  equation  of  a  straight  line. 

ART.  52.     A  circle  through  the  intersections  of  two  given 
circles. 

(Xz  +  y*  +  A*  +  B  y  +  C  =  o    (i)  ) 
f  \x*  +  f  +  AlX  +1^  +  ^=0(2)  (are  any  two  circles, 

then  (x2  +  y2  +  Ax  +  Ey  +  C)  + 

n(x2+y*  +  Alx  +  Bly  +  C1)  =  o    ...     (3) 


74  Analytical  Geometry. 

is  the  equation  of  a  circle  through  the  intersections  of  (i) 
and  (2).  For  since  (3)  is  a  combination  of  (i)  and  (2)  it 
must  contain  the  conditions  that  are  common  to  both,  and 
the  only  conditions  common  to  both,  in  general,  are  their 
points  of  intersection.  (3)  is  the  equation  to  a  circle,  for 
it  can  be  put  in  the  form, 

(i  +  n)  x*  +  (i  +  »)  y  +  (A  +  Aj»)  x  + 
(B+BX>:X+  (C  +  C1»)=o, 


or    S.+/+A  +  VS+    B+B,n          £±0,* 

i  -\-  n  i  +  w  i-f-w 


which  is  clearly  the  equation  to  a  circle  of  the  general  form. 
Further,  (3)  is  satisfied  by  any  point  that  satisfies  both 
(i)  and  (2).  for  (3)  is  made  up  exclusively  of  (i)  and  (2). 
If  a  third  condition  be  supplied,  n  can  be  determined  and 
a  definite  circle  through  (i)  and  (2)  results. 

EXERCISE. 
The  Circle. 

What  are  the  co-ordinates  of  the  centre  and  the  radii  of 
following  circles? 

1.  x2  +    y2  —  2  x  +  4  y  =  n. 

2.  x2  +  y2  —  6  y  =  o. 

3.  x2  +  y2  +  x  -  3  y  =  \f. 

4.  3  x2  +  3  y2  —  8  x  —  2  y  =  102  J. 

5.  x2  +  y2  +  8  x  =  33. 

6.  x2  +  y2  +  6  x  +  8  y  =  -  9. 

7.  4  x2  +  4  y2  -  2  #  +  y  =  -  TV 

8.  8  x2  +  8  y2  -  16  x  -  16  y  =  564. 


Analytical  Geometry,  75 

Write   the   equations   for   the   following  circles,    (h,   k) 
being  the  co-ordinates  of  the  centre,  and  r  the  radius. 
9.   h  =  -  2         k  =  3  r  =  4! 

10.  &  =  J  £  =  2j  r  =  4 

11.  h=  I  *--*  r=  V 

12.  h  =  o  k  =  i  r  =  5 

Find  the  equations  for  tangent  and  normal  to  following 
circles: 

13.  *2  +/=  9  at  (-_ij,  3). 

14.  *2  +  f  =  6  at  (v/2,  2). 

15.  *2  +  /  =  34  at  (-  3,  -  5). 

1 6.  x2  +  y2  =  25  at  point  whose  abscissa  is  3. 

17.  x2  +  ^2  =  1 6  at  point  whose  ordinate  is  —  x/T- 

18.  (*  +  2)2  +  0  -  i)2  =  100  at  (6,  7). 

19.  x2  +  (y-3)2=  25  at  (3,  ?). 

20.  #2  +  y2  =  20  at  (?,  2). 

Find  the  intersection  points  of  the  following: 

21.  #2  +  y2  =  25  and  x2  +  y2  +  14  x  +  13  =  o. 

22.  *2  +  y2  =  6  and  #2  +  /  =  8  x  -  8. 

23.  x2+y2-2x  —  4y  —  i  =  o, 
and  2  #2  -f  2  ;y2  —  8  #  —  12  y  +  10  =  o. 

24.  x2  +  y2  =  4,  and  ^c2  +  y2  +  2  #  —  3  =  o. 

25.  Find  the  equation  of  the  circle  passing  through  the 
intersections  of  .r2  -f  y2  =  9  and  3  x2  +  3  y2  —  6  x  +  8  y  =  i, 
which  also  passes  through  the  point  (4,  —  5). 

26.  Find  the  equation  of  the  circle  passing  through  the 
intersections     of    x2  +  y2  =  16     and    x2  +  ^2  +  2  rv  =  8, 
which  also  passes  through  the  point  (— i,  2). 

27.  Find  the  equation  of  the  circle  through  the  three 
points  (o,  o),  (2,  3),  and  (3,  4).     What  are  the  co-ordinates 
of  its  centre  and  its  radius  ? 

28.  Find  the  equation  of  the  circle  through  the  points 
(2,  ~  3),   (3,  -  4),  and  (-  2,  -  i). 


76  Analytical  Geometry. 

29.  Find  the  equation  of  the  circle  through  the  points 
(-  4,    -  4);   (-  4,  -  2);   (-  2,  +  2). 

30.  Find  the  equation  of  the  circle  passing  through  the 
origin  and  having  x  and  ^-intercepts  respectively  6  and  8. 

31.  Find  the  equation  of  a  circle  circumscribing  the  tri- 
angle whose  sides   are  x  +  2  y  =  o,   3  x  —  2  y  =  6,   and 

x  ~  y  =  5- 

32.  Find  the  equation  of  a  circle  passing  through  (i,  5) 
and  (4,  6)  and  having  its  centre  on  the  line  y  —  x  -\-  4  =  o. 

33.  Find  the  equation  of  a  circle  through   (3,  o)  and 
(2,  7)  whose  radius  is  5. 

34.  Find  the  equation  of  a  circle  having  the  line  joining 
(f ,  f )  to  the  origin  as  its  diameter. 

35.  Plot  by  points  the  circular  curve  whose  chord  is 
30'  and  sagitta,  9'. 


CHAPTER  VI. 
CONIC    SECTIONS. 

ART.  53.  The  sections  of  a  right  circular  cone  made  by 
a  plane  intersecting  it  at  varying  angles  with  its  axis,  are 
called  conic  sections. 

If  the  plane  is  parallel  to  an  element  of  the  cone  the 
intersection  is  called  a  parabola. 

If  the  plane  cuts  all  the  elements  of  one  nappe  of  the 
cone,  the  section  is  called  an  ellipse. 

When  the  plane  is  parallel  to  the  base  of  the  right  cone 
the  ellipse  becomes  a  circle. 

If  the  plane  cuts  both  nappes  of  the  cone,  the  section  is 
called  a  hyperbola. 

The  hyperbola  evidently  has  two  branches  (where  it 
intersects  the  two  nappes).  All  these  sections  are  called 
collectively  conies. 

ART.  54.     The  equation  of  a  conic. 

From  the  standpoint  of  analytical  geometry,  a  conic  is 
defined  as  a  curve,  the  distances  of  whose  points  from  a 
fixed  straight  line,  called  the  directrix,  and  from  a  fixed 
point,  called  the  focus,  bear  a  constant  ratio  to  each  other. 
This  ratio  is  called  the  eccentricity  of  the  conic.  It  can  be 
readily  proved  geometrically  that  this  definition  follows 
from  the  definitions  of  Art.  53. 

In  Fig.  28  let  P  be  any  point  on  a  conic,  the  ;y-axis  the 
directrix,  and  F  the  focus.  Draw  AP  perpendicular  to 
the  directrix,  PB  perpendicular  to  #-axis,  and  join  P  and 

PF 

F.     Call  the  constant  ratio  e,  then    - —  =  e, 

x  A. 

77 


Analytical  Geometry. 


or  PF  =  e.  PA (i 

The  co-ordinates  of  P  are  x  =  OB  -  AP,  y  =  PB. 
Represent  the  constant  distance  OF  by  p,  then 

PF2  =  FB2  +  PB2  (2)  [in  the  right  triangle  FPB]. 
FB  =  OB  -  OF  =  x  -  p.     PB  =  y. 

Substituting  in  (2);     PF2  =  (oc  -  p)2  +  y2. 


Hence   (i)  becomes,  \/(x  —  p)*  -\-  y*  =  ex. 
squaring;  (x—p)2jry2=e2x2, 

collecting;  (i  —  e2)x2+y2  —  2  px+j,'--=o  (a) 

which  is  the  equation  for  any  conic  in  rectangular  co-or- 
dinates. The  polar  equation  is  much  simpler.  It  may  be 
derived  by  transforming  (a)  to  polar  co-ordinates,  or  thus; 


Fig.  28. 

in  Fig.  28,  let  the  co-ordinates  of  P  be  p  =  PF,  6  =  Z  PFB, 
the  pole  being  at  F  and  the  #-axis  being  the  initial  line. 

Then    cos  PFB  =     —  ,  or  FB  =  FP  cos  PFB  -  p  cos  0. 

But  FB=OB-OF=AP-  OF  =  AP-& 

that  is,  p  cos  0  =  AP  -  p, 

whence  AP  =  p  cos  6  +  p. 


Analytical  Geometry. 

Substituting  in  (i);  p  =  e  (p  cos  6  +  p)  =  ep  cos  6  + 
Transposing  and  collecting; 

p  (i  —  e  cos  6}  =  ep. 


79 


i  —  e  cos  0 

THE    PARABOLA. 

ART.  55.  The  parabola  is  defined  in  analytical  geom- 
etry as  a  curve,  every  point  of  which  is  equally  distant  jrom 
a  fxed  point  and  a  fixed  straight  line.  This  definition  is 
in  entire  accord  with  Art.  53.  Y  / 

Clearly  from  this  definition  A 
e  —  i  in  the  parabola,  hence  (a) 
becomes  y2  --  2  px  -f  p2  =  o, 
or  y2  =  2  px  —  p2  (i).  As  it 
is  usually  convenient  to  have 
the  origin  at  the  vertex  O  (in 
Fig.  29)  of  the  parabola,  and 
as  the  vertex  is  midway  between 

the  directrix  and  the  focus  by  definition,  the  above  equa- 
tion is  transformed  to  new  axes  having  their  origin  at  the 

vertex  by  substituting  (xf  +  *  j  for  x  and  leaving  y  un- 
changed. 

The   co-ordinates    of   the  new  origin    are  f*-i  oi  with 


c 

7 

P 
B 

0 

vj 

TJ 

F 
V 

respect  to  the  old,  hence  the  transformation  equations  are 
as  above, 

x  =  x'  +  i-  and  y  =  /; 

2 

(i)  then  becomes  y'2  =  2  p  (x*  +  *)  —  p2  =  2  pyf, 

or     [dropping  accents]       y2  =  2  px (A^) 

The  equation  is  derived  directly  from  the  definition,  thus: 


8o  Analytical  Geometry. 

In  Fig.  29,  let  P  be  any  point  on  the  parabola;  AC,  the 
directrix,  O  the  vertex  and  the  origin.  Draw  AP  ||  and 
PB  perpendicular  to  the  #-axis,  and  let  F  be  the  focus. 

Then  if  DF  be  represented  by  p,  OF  will  equal  £-  by  defi- 

2 

nition. 

PF  =  PA     .....     (a)  [JDV  definition  of  parabola] 


But    PF  =  xPB*  +  FB=  -vPB*  +  (°B  ~  OF) 


and        PA  =  OB  +  DO  =  x  +      . 

2 


2 


Substituting  in  (a);  i /y*  _^_  /x  __  PY  «#-}-£, 
V  \         2  / 

I          -b  \2       /         ^?  \2 
squaring;  /  -f  (x  —    f-  }   =  lx  +  L  \ 

\  2  /         \         2  I 


y2  —  2  px,  as  before. 

From  its  equation,  the  characteristic  property  of  a  para- 
bola is,  that  the  ratio  of  the  square  of  the  ordinate  of  any 
point  on  it  to  the  abscissa  of  that  point  is  a  constant,  for 

J—  =  2  p.    This  relation  is  used  in  physics  to  show  that  the 

OC 

path  of  a  projectile  is  a  parabola.  When  the  curve  is 
symmetrical  to  the  ;y-axis  as  in  Fig.  30,  the  equation  takes 
the  form,  x2  =  2  py. 

As  an  exercise  prove  this  last  equation. 

ART.  56.  If  in  the  equation  to  the  parabola  (Ap),  the 
abscissa  of  the  focus  (F),  x  =  *  be  substituted,  the 


Analytical  Geometry. 


81 


resulting  values   of  y  are   the   ordinates  of   the  points  on 
the  parabola  immediately  over  and  under  the  focus;    ; 


thus  y2  =  2 

whence  y  =  ±  p. 

These  two  ordinates  together,  extending  from  the  point 


Fig.  30. 

above  the  focus  to  the  point  below  on  the  curve,  form  what 
is  called  the  latus  rectum.  (GH,  Fig.  29.) 

The  latus  rectum  evidently  equals  2  p,  and  is  often  called 
the  double  ordinate  through  the  focus. 

ART.  57.     To  construct  the  parabola. 

First  Method.  The  definition  suggests  a  simple  mechan- 
ical means  of  constructing  the  parabola.  Let  the  edge  of 
a  T-square  (AB,  Fig.  31)  represent  the  directrix;  adjust  a 
triangle  to  it,  with  its  other  edge  on  the  axis,  as  DEC. 
Attach  one  end  of  a  string  whose  length  is  EC,  at  C  and 
the  other  end  at  F.  Keeping  the  string  taut  against  the 


82 


Analytical  Geometry. 


Fig.  3i. 


base  of  the  triangle  with  a  pen- 
cil (as  at  G)  slide  the  ruler 
along  the  T-square  and  the 
point  of  the  pencil  will  de- 
scribe a  parabola,  for  every- 
where it  will  be  equally 
distant  from  AB  and  F, 
as  at  G;  for  EG  =--  GF, 
since  GF  =  =  E'C'  -  GC' 
=  EC  --  GC'  and  E'G 
=  E'C'  -  GC'. 

Second  Method:  For  practical  purposes  it  is  more  con- 
venient to  construct  by  points. 

Let  AB  (Fig.  32)  be  the  directrix;  F,  the  focus,  and  OX, 
the  axis.  Lay  off  as  many  points  as  desired  on  the  axis, 
as  C,  D,  E,  G,  H,  etc.;  then  with  F  as  a  centre  and  radii 
successively  equal  to  OC,  OD,  OE,  OG,  OH,  etc.,  draw 
arcs  above  and  below  OX,  at  C,  D,  E,  G,  H,  etc.;  erect 
perpendiculars  to  OX  in- 
tersecting these  arcs  at 
C'  and  C",  D'  and  D", 
E'  and  E",  etc. 

These  points  of  inter- 
section will  be  points  on 
the  parabola,  for  they 
are  all  equally  distant 
from  AB  and  F  by  the 
construction. 

By  taking  these  points 
sufficiently  near  together, 
the  parabola  can  be  constructed  as  accurately  as  desired. 
ART.  58.  The  polar  equation  to  the  parabola  is  easily 
derived  from  the  general  polar  equation  to  a  conic,  by 
remembering  that  for  a  parabola,  e  =  i. 


Analytical  Geometry.  83 


Hence  r  — 


i  —  ecos 

A 

becomes  p  = 


i  —  cos  0 

ART.  59.  It  is  evident  from  the  form  of  the  parabola 
equation,  y2  =  2  px,  that  x  cannot  be  negative  without 
making  y  imaginary,  hence  no  point  on  the  parabola 
y2  =  2  px  can  lie  to  the  left  of  the  Y-axis;  that  is,  the  curve 
has  but  one  branch  lying  to  the  right  of  the  Y-axis.  In 
order  to  represent  a  parabola  lying  to  the  left  of  the  origin, 
the  equation  would  have  to  take  the  form 

/  =    -    2  px, 

so  that  negative  values  of  x  would  make  y2-  positive. 
In  this  latter  case  no  positive  value  of  x  would  satisfy. 

EXERCISE. 

What  are  the  equations  of  the  parabolas  passing  through 
the  following  points,  and  what  is  the  latus  rectum  in  each 
case? 

I.    (1,4);         2.    (2,  3);         3.    (i  J);        4-    (3,-4). 

5.  The    equation    of    a    parabola    is    y2  =  4  x.     What 
abscissa  corresponds  to  the  ordinate  7  ? 

6.  What  is  the  equation  of  the  chord  of  the  parabola 
y-  =  8  x,  which  passess  through  the  vertex  and  the  nega- 
tiv-e  end  of  the  latus  rectum? 

7.  In  the  parabola  y2  =  9  x,  what  ordinate  corresponds 
to  the  abscissa  4?     Construct  the  following  parabolas. 

8.  y2  =  6  x.  9.      x2  =  9  y. 
10.   y''  =    -  4cc.                               ii.     x2  =  —Sy. 

12.  For    what    points    on    the    parabola    y2  =  8  x    will 
ordinate  and  abscissa  be  equal  ? 

13.  What    are   the   co-ordinates   of   the   points   on   the 


84 


Analytical  Geometry. 


parabola  y2  —  10  x,   if  the    abscissa  equals   f  of  the  or- 
dinate  ? 

Find  intersection  points  of  the  following: 

14.  y2  =    4X  and  y   =  2^—5. 

15.  y2  =  18  x  and  y   =  2^—5. 

16.  y2  =      4X  and  #2  +  y2  ==  12. 

17.  y2  =  16  ^  and  #2  +  ^2  —  8  #  =  33. 

1 8.  What   does   the   equation  y1  =  2  px  become  when 
the  origin  is  moved  back  along  the  axis  to  the  directrix  ? 

ART.  60.     To  find  the  equation  of  a  tangent  to  the  para- 
bola. 

The  process  employed  to  find  the  equation  of  a  tangent 
to  the  circle  is  just  as  effective  for  the  parabola. 

If  in  the  equation  to 
a  line  through  two  given 
points,  the  points  be 
situated  on  a  parabola, 
and  hence  are  deter- 
mined by  its  equation, 
X  the  equation  becomes 
that  of  a  secant  to  the 
parabola.  If  the  two 
points  are  then  made  to 
approach  coincidence, 
the  secant  becomes  a 


Fig.  33- 


In  the  equation  to  a  straight  line, 


tangent. 


(B) 


let  the  points   (x',  y')  and    (x",  y"}  be  on  the  parabola 
y2  =  2  pxi  then  the  two  equations  of  condition 

y'2  =  2px' (2) 

y2  =  2  px" (3) 


Analytical  Geometry.  85 

arise    from    substituting    these    values    in    the    parabola 
equation. 

Subtracting  (2)  from  (3); 

yn  _  yn  =  2  py*  -  2px'=  2p  (x"  -  x'). 

Factoring;     (/'  -  /)  (/'  +  /)  =  2  p  (x"  -  x'). 

Dividing  through  by     (/'  +  /)  (x"  —  *'), 

x"  -xf==  y"  +'/' 


Substituting  this  value  of  the  slope    -^—-^7,   in    (B); 

x    —  x 


y  —  yf  =  P  f  (x  —  x'}  (4),  which  is  now  the  equa- 
tion of  a  secant  line  to  the  parabola,  say  ABC  (Fig.  33), 
the  point  B  being  (x",  y")  and  C  being  (xf,  y'). 

If  now  the  point  B  approach  C,  (x",  yf)  approaches 
(xf,  /)  and  eventually  x"  =  x'  and  y  "  =  /,  and  the  secant 
ABC  becomes  the  tangent  DCE. 

Making  x"  =  x',  y"  =  /  in  (4),  it  becomes, 


7  -y  -  ,  c*  -X) (TP) 

which  is  the  equation  to  the  tangent  DCE  at  the  point 

(*',/)• 
Simplifying  (Tp),  yyr  -  y'2  =  px  -  pxf 

yyf  —  2  pxf  =  px  —  px'  [since  y'2  =  2  px']; 
or      yy'  =  p  (x  +  xf)   (Tp')  [transposing,  collecting  and 
factoring]. 

Corollary:  The  tangent  intercept  on  the  X-axis,  OD,  is 
found  by  setting  y  =  o  in  (Tp). 
Whence  o  =  p  (x  +  yf\ 

x=  -  x'. 


86  Analytical  Geometry. 

That  is,  the  intercept  is  equal  to  the  abscissa  of  the  point 
of  tangency,  with  opposite  sign. 

ART.  61.     The  equation  to  the  normal. 

Since  the  normal  is  perpendicular  to  the  tangent  through 
the  same  point,  it  has  the  same  equation  except  for  its 
slope,  which  is  given  by  the  relation  for  perpendicular  lines, 


m 


In  the  tangent  equation  m  =    ^ 
Hence  the  normal  equation  is 


In  Fig.  33,  CG  is  the  normal  at  C. 

ART.  62.  The  equation  of  the  tangent  in  terms  of  its 
slope. 

As  in  the  case  of  the  circle  it  is  only  necessary  to  deter- 
mine the  constants  in  the  slope  equation  of  a  straight  line, 
so  that  it  has  but  one  point  in  common  with  the  parabola. 

The  equations  to  parabola  and  line  are, 

f  =  2  px     ........     (i) 

and  y  =  mx  +  b     .......     (2) 

Eliminating  y,  to  find  the  intersection  equation  for  x, 

(mx  +  b)2  =  2  pXj 

m2x2  +  2  mbx  +  b2  =  2  px, 

m2x2  +  (2  mb  -  2  p)  x  +  b2  =  o      .     .     (3) 

The  two  values  of  x  in  equation  (3)  will  be  the  abscissas 
of  the  two  points  of  intersection.  These  two  points  will 
coincide  if  the  two  values  of  x  are  the  same,  and  this  can 


Analytical  Geometry.  87 

only    occur    if    m2x2  -f  (2  mb  —  2  p)  oc  +  b2    is    a    perfect 
square. 
By  the  binomial  theorem  this  is  the  case,  if 

x2  (mb  —  py  =  m2x2b2 
or  m2b2  -  2  pmb  +  p2  = 

whence          2 


2  w 
Substituting  this  value  of  b  in  (2), 


which  is  the  equation  of  the  tangent  in  terms  of  its  slope. 

ART.  63.  Equation  to  the  normal  in  terms  of  the  slope 
of  the  tangent. 

Combining  (Tm?p)  with  the  equation  to  the  parabola, 
we  get  the  co-ordinates  of  the  point  of  tangency  in  terms  of 
m  and  p.  Since  the  normal  passes  through  this  point  it  is 
necessary  to  know  these  co-ordinates. 

Combining  then,  y'2  =  2  pxf 

and  y  =  mxf  +    —  , 


2  m 


we  get  off  =  ~—2  ,y'=—  IX,  /  being  point  of  tangency]. 

The  slope  of  the  normal  is  m'  =  -    -  [since  it  is  perpen- 

m 

dicular  to  the  tangent,  whose  slope  is  m]. 

The  equation  to  a  line  through  a  given  point  with  a 
given  slope,  m',  is  y  —  yf  —  m'  (x  —  x'}  .....  (C) 
Substituting  in  (C)  values  of  x',  y',  and  w', 


m  m  2  m' 

+  m2x  =  pm2  +    £•  . 


88 


Analytical  Geometry. 


This  equation  being  a  cubic  in  w,  three  values  of  m  will 
satisfy  it,  hence  through  any  point  on  the  parabola  three 
normals  can  be  drawn,  having  the  three  slopes  given  by 
the  three  values  of  m. 

ART.  64.  The  following  property  of  a  parabola  has  led 
to  its  application  for  reflectors,  making  it  of  peculiar  in- 
terest in  optics. 

To  show  that  the  tangent  to  the  parabola  makes  equal 
angles  with  a  line  from  the  focus  to  the  point  of  tangency 


G  0 


-K 


FT  R 


Fig.  34- 

(a  focal  line),  and  a  line  drawn  through  the  same  point 
parallel  to  the  axis  of  the  parabola. 

LM  (Fig.  34)  is  a  tangent  to  the  parabola  PON  at  P, 
intersecting  the  axis  produced  at  L. 

Draw  the  focal  line  FP  and  PK  ||  to  the  axis  OX.  Then 
ZLPF=  ZMPK. 

By  Art.  60,  Cor.,  the  tangent  ^-intercept,  OL  =  —  yf 
,  /)  being  point  of  tangency,  P]. 


Analytical  Geometry.  89 

Also  OF  =  £   [by  structure  of  the  parabola]. 

2 

/.  LF  =  xr  +  2.  [the  sign  of  yf  is  neglected  for  we 
want  only  absolute  length]. 

Let  QS  be  the  directrix.     Then 

PF  =  PQ  =  GT  =  GO  +  OT  =  £  +  x'.     [OT  =*'.] 

.;,  LF  =  PF,  and  triangle  LPF  is  isosceles; 
hence  Z  LPF  =  Z  PLF. 

But  Z  PLF  =  Z  MPK    [since  PK  is  ||  to  LX]. 

.-.       Z  LPF  =  Z  MPK. 

Let  PR  be  the  normal;  then  Z  FPR  =  Z  RPK 
[since  Z  LPF  =  Z  MPK,  and  LPR  =  MPR,  being  right 
angles]. 

Since  the  angles  of  incidence  and  reflection  are  always 
equal  for  light  reflected  from  any  surface,  it  follows  that 
light  issuing  from  a  source  at  F  would  be  reflected  from  the 
surface  of  a  paraboloid  mirror  in  parallel  lines,  (as  PK). 

ART.  65.  The  diameter  of  any  conic  may  be  defined  as 
the  locus  of  the  middle  points  of  any  series  of  parallel 
chords. 

A  chord  is  understood  to  be  a  straight  line  joining  any 
two  points  on  the  curve.  In  Fig.  35,  AB  being  the  locus 
of  the  middle  points  of  the  system  of  parallel  chords,  of 
which  CD  is  one,  is  a  diameter  of  the  parabola  PON. 

ART.  66.  To  find  the  equation  oj  a  diameter  in  terms  of 
the  slope  of  its  system  of  parallel  chords. 


9o 


Analytical  Geometry. 


Draw  (Fig.  35)  a  series  of  chords  (like  CD)  ||  to  each 
other.  To  determine  the  locus  of  the  middle  points  of 
these  chords,  that  is,  the  diameter  corresponding  to  them. 

Let  the  equation  of  any  one  of  the  chords,  as  CD,  be 


and 


y  =  mx  +  b       (i), 
y2  =  2  px  (2)  be  the  parabola  equation. 


If  (i)  and  (2*)  be  combined  as  simultaneous,  the  co-ordi- 
nates of  C  and  D,  the  points  of  intersection,  will  be  found. 
First  to  find  the  abscissa,  eliminating  y  by  substituting ; 


Fig.  35. 

(mx  +  b)2  =  2  px, 
m2x2  +  2  mbx  +  b2  =  2  pxt 

"*V*U+4=    o      .          (3) 


Now  in  a  quadratic  of  the  form  z2  +  az  +  b  =  o,  the  sum 


Analytical  Geometry.  91 

of  the  two  values  of  the  unknown  equals  the  coefficient  (a) 
of  the  first  power  of  the  unknown  with  its  sign  changed.* 

Hence  the  two  values  of  x  in  (3),  which  are  the  abscissas 
respectively  of  C  and  D,  added  together,  equal  the  coeffi- 
cient of  x  in  (3)  with  its  sign  changed. 

Call  the  co-ordinates  of  C  and  D  respectively  (xf,  y') 
and  (x?t  /')• 

,    ,  2  nib  —  2  p 

Then  x'  +  oc"  = 


m 
Eliminating  x  from  (i)  and  (2),  we  get  from  (i) 


m 


Substituting  in  (2);  y*  =  2  py  ~  2 

m 


f-^L+trL.  =  0     .     .    .    .     (4) 
m  m 


by  principle  cited  above,    y'  +  y"  — 


m 


In  Art.   32   it  was  shown  that  the  co-ordinates  of  the 
middle  point  of  a  line  joining  (x*,  /)  and  (V',  /')  are, 

/*'+;*"         /+/\ 


and 


but       -^  +  Va2-4^.         ~  a  ~  Xa2  -  4 

2  2 

coefBcient  of  z  with  its  sign  changed. 


92  Analytical  Geometry. 

Calling  the  co-ordinates  of  the  middle  point  (E)  of  CD, 
(X,  Y). 

TU  v        *?  +  xff.  mb  —  p  f  . 

Then  X=     _  _____    ...     (5) 

and  Y  =  -^±^=-2  ......     (6) 

2  m 

Remembering  that  an  equation  to  a  line  must  express 
a  constant  relation  between  the  co-ordinates  of  every  point 
on  that  line,  it  is  clear  that  b  cannot  form  a  part  of  the  equa- 
tion we  are  seeking,  for  b,  the  y-intercept,  of  the  chords, 
is  different  for  every  chord,  but  m  is  constant,  since  the 
chords  are  all  parallel.  It  would  ordinarily  be  necessary 
then  to  eliminate  b  between  (5)  and  (6),  but  in  this  case 
(6)  does  not  contain  b  and  hence  it  represents  the  true 
equation  for  the  diameter.  We  will  designate  it  thus  : 


It  evidently  represents  every  point  on  this  diameter,  for 
CD  was  any  chord,  and  hence  the  expression  for  its  middle 
point  will  apply  equally  well  to  all  the  chords. 

Cor.  I  :  The  form  of  this  equation  shows  that  the  diam- 
eter is  always  parallel  to  the  X-axis,  that  is,  to  the  axis  of 
the  parabola. 

Cor.  II  :  Combining  (Dp)  with  the    parabola  equation, 
we  get  the  co-ordinates  of  their  point  of  intersection,  (A). 
y2  =  2  px, 

y  =  t 

m 
whence  *-r  =  2  px 


2m*          tn 


Analytical  Geometry.  93 

By  Art.  63  it  was  found  that  the  tangent  whose  slope  is 

m  touches  the  parabola  at  the  point  [  -£— ,   £•),]  which  is  A 

\2  m2    m    I 

here.  Hence  in  this  case  the  tangent  at  A  has  the  same 
slope,  m,  as  the  parallel  chords,  and  is,  therefore,  ||  to  them. 

That  is,  the  tangent  at  the  end  of  a  diameter  is  parallel  to 
its  system  of  parallel  chords. 

Definition:  The  chord  that  passes  through  the  focus  is 
called  the  parameter  of  its  diameter. 

ART.  67.  The  two  following  propositions  are  interesting 
as  applications  of  the  principles  already  discussed. 

To  find  the  equation  to  the  locus  of  the  intersection  of 
tangents  perpendicular  to  each  other. 

It  is  plainly  necessary  to  find  the  concordant  equations 
of  any  two  perpendicular  tangents  and  by  combining  their 
equations  get  their  intersection  point. 

The  slope  equation  for  any  tangent  is 

then  y  =  m'x  -\ £—  ,   (2)  will  represent  any  other  tangent. 

2  m' 

If   the  two  tangents  are  perpendicular  to  each  other  then 

m'  =  —   -  ,  and  (2)  becomes,  y  =  —  —  —  ^    .     .     (3) 
m  m        2 

Subtracting  (3)  from  (i), 


o  = 


=  (  m  H  —  )  oo  +  —(  m  -\  —  ];  whence  x  =  —  —  • 
\  m  )  2  \  m]  2 

This  equation   being  the   combination   of    (i)   and    (3) 
represents  their  intersection,  that  is,  it  is  the  equation  of 

the  locus  of  all  intersections.     But   x  =  —  £  is  the  equa- 

2 

tion  of  the  directrix,   hence  all  tangents  to  the  parabola 


94  Analytical  Geometry. 

that   are   perpendicular    to    each    other    intersect    on    the 
directrix. 

ART.  68.  To  find  the  locus  of  the  intersection  of  any  tan- 
gent, with  the  perpendicular  upon  it  from  the  focus. 

The  equation  of  any  tangent  line  is  y  =  mx  H &-  ,  (i). 

The  equation  to  a  line  through  the  focus  having  the  slope 
mr  is  by  (C),  y  =  in'  lx  —  *-  ),  (2).     The   focus  being  the 

point  [£-,  o)    .   Since  (2)  is  perpendicular  to  (T),  m'  =        — , 

hence  (2)  becomes  y  = -lx  —  £-),  or  y=  -     -  +  -£-,  (3). 

;;/  \         2  /  m      2  w 

Subtracting  (3)  from  (i),     o  =  ( j  H )  x. 

\         m/ 

Whence  x  =  o, 

But  x  =  o  is  the  equation  of  the  Y-axis,  .'.  every  tangent 
to  the  parabola  intersects  the  perpendicular  upon  it  from 
the  focus  on  the  Y-axis. 

ART.  69.  It  is  sometimes  desirable  to  express  the 
equation  of  a  parabola  with  reference  to  a  point  of  tangency 
as  origin,  and  with  the  tangent  and  a  diameter  through 
the  point  of  tangency  as  axes. 

Knowing  the  co-ordinates  of  the  point  of  tangency  in 
terms  of  the  tangent  slope  and  knowing  that  the  diameter 
is  ||  to  the  axis,  it  is  easy  to  apply  the  transformation 
equations  in  Art.  38. 

Remembering  that  the  new  X-axis  (a  diameter)  is  parallel 
to  the  old,  hence  6=0,  and  that  tan  cjy  =  m,  since  the 
new  Y-axis  is  a  tangent  and  (j)  is  the  angle  it  makes  with 
the  old  X-axis. 


Analytical  Geometry.  95 

Also  (a,  b)  the  co-ordinates  of  the  new  origin  become, 


(  x  =  a  +  x/  cos  6  -f  y  cos  <k. 

Equations <  t         /    •     /i    ,     /    •      i 

J  ^y  =  &  -f  x  sin  o'  +  7  sm  9, 


become,  #  =  — *• 1-  xf  +  y  cos  <^ 

2  w2 

[since  cos  6  =  cos  o  =  i]. 
y  =  £  +  y  sin  0 

[since  sin  0  =  sin  ^  =  o]. 
Substituting  in  the  parabola  equation, 


we  get, 


f=2 


sn (h     =  2  p     -—  +  xf  +  /  cos 
2  m2 


sin  (h 

=   r- 


or  since  m  =  tan 

cos 

(p  cos  <j>    .      ,     • 
sin  c£ 


+  2  ^T^C«^>  +  ;y/2  sin2 


sin2  0=2 


Since  we2  0  =  cot2  c6  +  i  =  -^—  + 


this  may  be  written,    y2  =  — £   x  +  2  px, 

m2 

y*=*p(*+™>)x> 


96  Analytical  Geometry. 

where  m  is  the  tangent's  slope,  or  the  tangent  of  the 
angle  it  makes  with  the  axis  of  the  parabola. 

ART.  70.  The  parabola  is  of  practical  interest  also  in 
its  application  to  trajectories. 

By  the  laws  of  physics  a  projected  body  describes  a 
path,  determined  by  the  resultant  of  the  forces  of  projec- 
tion and  of  gravity  acting  together  upon  the  moving  body 
[neglecting  air  resistance]. 

In  a  given  time,  t,  with  a  velocity,  v,  a  body  will  move  a 
space,  s=  vt.  (i).  Meanwhile  it  falls  through  a  space 

S  =  —  gt2.    (2)  [g  =  acceleration  by  gravity.] 

2 

Square  (i)  and  divide  by  (2) 


S         g 

It  is  easy  to  see  that  the  horizontal  distance,  s,  which 
the  body  wrould  move  if  undiverted  by  gravity,  is  like  an 
abscissa,  and  that  the  vertical  space,  S,  that  the  body 
would  fall  by  action  of  gravity,  is  like  an  ordinate. 

Also       — :        is  clearly  a  constant,  (like  2  p). 

g 

*>  9  2 

Hence    -=  ^—  or  s2  =  ^-  S  is  exactly  like  y2  =  2  px. 
S       g  g 

That  is,  the  path  of  a  projectile  is  a  parabola,  if  we  neglect 
the  resistance  of  the  air. 

EXERCISE. 

Find  the  equations  of  the  tangents  to  each  of  the  follow- 
ing parabolas : 

1.  f  =  6x         at  (§,  4). 

2.  y2  =  9  x         at  (4,  6). 

3.  x2  =  6y         at  (6,  6). 


Analytical  Geometry.  97 


4-  y*=  -4* 

5.   /  =  4  a# 

at  (-  i,  2) 
at  (*',  /). 
at  (4^,  ?). 

s:  J:r4r 

at  (?-4). 
at  (6,  ?). 

9.  Find  the  equation  of  the  normal  to  each  of  the  pre- 
ceding parabolas. 

10.  Find  the  equations  of  the  tangents  to  the  parabola 
y2  —  8  x  from  the  exterior  point  (i,  3). 

11.  Find  the  equation  of  the  tangent  to  y2  —  9  x  par- 
allel to  the  line  2^=3^—5. 

12.  Find   the  equation  of  the  tangent  to  the  parabola 
•y2  =  4  x  perpendicular  to  the  line  y  +  3  x  =  i. 

13.  Find  the  slope  equation  of  the  tangent  to  the  para- 
bola x2  =  2  py. 

14.  Find   the  equation  of  the  tangent  to  the  parabola 
y2  =  8  x  from  the  point  (i,  4). 

15.  Find  the  equation  to  the  tangent  at  the  lower  end  of 
the  latus  rectum. 

16.  The  equation  to  a  chord  of  the  parabola  y2  =  4  x 
is  5  y  ~~  2  x  ~   I2  =  °-     What    is    the    equation  of  the 
diameter  bisecting  it  ? 

17.  What   is  the  equation  of  the  parabola  referred  to 
this  diameler  and  the  tangent  at  its  extremity? 

18.  In  the  parabola  y2  =  8  x,  what  is  the  parameter  of 
the  diameter  whose  equation  is  y  =  16? 

19.  What    is    the    equation    of   the   parabola   to   which 
2y=3#+8is  tangent  ? 

20.  The  equation  of  a  tangent  to  the  parabola  y2  =  9  x 
is  3  y  —  x=  ii.     What  is  the  equation  of  the  diameter 
through  the  point  of  tangency? 

21.  What  is  the  equation  to  the  chord  of  the  parabola 
^2  _  5  X)  which  is  bisected  at  the  point  (3,  4)  ? 


98  Analytical  Geometry. 

22.  The  base  of  a  triangle  is  10  and  the  sum  of  the 
tangents  of  the  base  angles  is  2.     Show  that  the  locus  of 
the  vertex  is  a  parabola  and  find  its  equation. 

23.  The  equation  to  a  diameter  of  the  parabola  y1  =9  x, 
is  y  =  —  3.     Find  the  equation  of  its  parameter. 

24.  Find  the  equation  of  the  diameter  to  the  parabola 
x2  =  2  py. 


CHAPTER  VII. 
THE    ELLIPSE. 

ART.  71.  The  ellipse  is  defined,  for  the  purposes  of 
analytics,  as  a  curve  every  point  of  which  has  the  sum  of 
its  distances  from  two  fixed  points,  called  foci,  always  the 
same;  that  is,  constant.  It  will  be  seen  later  that  it  is  a 
conic  in  which  e  <  i. 

The  line  AA'  (Fig.  36),  through  the  foci,  F  and  F',  ter- 


minated by  the  curve  is  called  the  major  or  transverse 
axis:  the  line  BB'  perpendicular  to  AA'  at  its  middle 
point  and  terminated  by  the  curve,  is  called  the  minor  or 
conjugate  axis. 

ART.  72.  To  find  the  equation  of  the  ellipse,  taking  the 
centre  O  (Fig.  36)  as  origin  and  the  major  and  minor 
axes  as  co-ordinates  axes.  Draw  PF'  and  PF,  lines  from 
any  point,  P,  to  the  foci  (focal  lines). 

Also  PD  perpendicular  to  AA'. 

Call  the  co-ordinates  of  P,  (x,  y)  [(OD,  PD)  in  Fig.  36] 

99 


ioo  Analytical  Geometry. 

represent  J  AA'  =  OA,  by  a;  J  BB'  =  OB,  by  b,  PF,  by 
;•;  PF',  by  /;  OF  =  J  FF',  by  c. 

It  is  required  to  find  the  relation  between  PD  and  OD, 
using  the  constants,  a,  b,  and  c.  The  right  triangles  PDF 
and  PDF',  immediately  suggest  the  means,  as  they  contain 
together  the  co-ordinates  (x,  y)  and  part  of  the  constants, 
and  also  PF  and  PF'  whose  sum  is  a  constant  by  definition. 

In  PDF,      PF2  =  PD2  +  DF2, 

or  r2  =  y2  +  (c  -  x)2, 


r  =  Vy~  +  (c  ~  x)2  ......     (i) 

In  PDF'     PF'2  =  PD2  +  DF'2, 

or  V2=  y2  +  (c  +  x)2 


or  r'  =  V/  +  (c  +  xr  -     .....     (2) 

By  definition  r  +  /  =  a  constant;  let  us  try  to  deter- 
mine this  constant.  Since  the  points  A  and  A'  are  on  the 
ellipse  they  must  obey  this  definition;  hence  FA  +  F'A  = 
this  constant. 

But  F'A  +     FA  =  FF'  +  2  FA. 

Also  F'A  +    FA  =  F'A'  +  FA'  =  2  F'A'  +  F'F. 

That  is,       5^  +  2  FA  =  2  F'A'  +  £#, 
whence  FA  =  F'A'. 

/.  FA  +  F'A  =  F'F'  +  2  FA  =  F'F  +  FA  +  F'A'  =  2  a. 

/.  r  +  /  =  2  a. 

Adding  (i)  and  (2); 


+  (c  -    x)2  +  yy2  +  (c  +  x)2  =  r  +  /  =  2  a    (3) 
Transposing  and  squaring; 

y*  +  (c  +  x)2  =  4  a2  -  4  a  vY  +  (c  -  x)2  +  y2 

+  (c-x)2 


j-r2cx      2T=4a   —  4  a  \x 

+  /   +/-    2  CX  +./ 

whence       —  4  ex  -{  4  a2  =  4  a  \/y2  +  (c  —  x)2. 


Analytical  Geometry.  101 

Dividing  by  4  and  squaring  again; 

c2x2  -  2j^x  -f  a4  =  ay  +  a2  c2-2^oHoc+a2x2 

ay  +  (a2  -  c2)  *2  =  a2  (a2  -  c2) (4) 

The  form  of  this  equation  may  be  readily  changed  by 
expressing  c  in  terms  of  a  and  b. 
The  point  B  being  on  the  ellipse, 

BF  +  BF'  =  2  a, 

but  BF  =  BF'  (since  BB'  is  perpendicular  to  AA'  at  its 
middle). 

BF=  a. 
In  the  right  triangle  BOF, 

BF2  =    5^2    +   Qf  2  =  s    £2    +  ^'^   ^  ^' 

that  is,  a2  =  b2  +  c2    '•  >J  * *'»   c    \       • ,-  'V  \ ;-'  -  «.' 

or  b2  =  a2  —  c2. 

Substituting  in  (4) 

a2?2  +  62*2  =  a262 (Ae) 

The  form  of  this  equation  shows  that  the  curve  is  sym- 
metrical with  respect  to  its  two  axes. 

Corollary:  The  polar  equation  to  the  ellipse  is  that  of 
the  conic  in  general, 

=  eP 

i  —  e  cos  6 ' 

where  p  =  distance  from  directrix  to  focus  and  e  <  i. 

ART.  73.  There  are,  by  definition,  two  latera  recta,  one 
through  each  focus.  Since  they  are  ordinates,  their  values 
are  found  by  substituting  in  the  equation  the  abscissas  of 
the  foci,  that  is,  x  =  ±  c  =  ±  \/a2—  b2. 

Substituting  this  value  of  x  in  (Ae), 

a2/  +  b2  (a2  -  b2)  =  a2b2, 

whence  y2  =  -      y  =  ±  —  • 

a2  a 

That  is,  2  y  =  latus  rectum  =    * —  • 

a 


102 


Analytical  Geometry. 


ART.  74.     To  find  the  -value  of  p  in  the  ellipse. 

In  Fig.  37,  NF'  =  p  in  general  equation  to  a  conic. 

A'F' 
Also      — —  =  e,  since  A'  is  a  point  on  the  conic  A'B  AB' 

(the  ellipse),  whence  A'F'  =  e  A'N (i) 

Also  AF' =  eAN,  (2).     [Since  A  is  a  point  on  conic.] 

Add  (i)  and  (2); 
A'F'  +  AF'  =  e  (A'N  +  AN)  =  e  (A'N  +  A'N  +  AA') 

or  AA'  =  e  (2  A'N  +  2  A'O)  =  2.1  (A'N  +  A'O)  =  2  e  ON, 


that  is>  2  a  =r  2  e  ON  -cfr  ON  =  - 


Again,  NF'  =  NO  -  OF'  =   -  -  c  =   -  -  ae; 


(3) 


Subtract  (i)ffom  (2);" 

AF'  -  A'F'  =  e  (AN  -  A'N)  =  e  AA'  =  2  ae. 
But  AF'  -  A'F'  =  AF'  -  FA 

[since  FA  =  A'F',  Art.  82]  =  FF  =  2  c. 
2  ae  =  2  c  [since  FF'  =  2  c\ 

c  =  ae   .  ...     (4) 


that  is,      NF'  =  />  = 


Analytical  Geometry. 


103 


Hence  the  polar  equation  to  the  ellipse  may  be  written, 


P  = 


a    i- 


i  —  e  cos  6 


=  [taking  F'  as  pole]. 
Va*  -  b2 


Also  from  (4)     e  =  - 

a  a 

Since  c  <  a,  e  is  always  less  than  i,  by  above  equation. 

This  is  expressed  thus;  the  eccentricity  of  the  ellipse  is 
the  ratio  between  its  semi-focal  distance  and  the  semi- 
major  axis. 

ART.  75.  The  sum  of  the  focal  distances  of  any  point  on 
the  ellipse  equals  the  major  axis. 

We  know  by  the  definition  of  the  ellipse  that  this  sum  is 
a  constant;  now  we  will  show  that  this  constant  is  the 
major  axis  from  its  equation. 

Let  P  be  any  point  on  the  ellipse  ABA'B'.     (Fig.  38.) 

Draw  the  focal  radii  FT  and  FP,  also  PD  perpendicular 
to  AA',  the  major  axis. 

The  co-ordinates  of  P  are   (OD,  PD),  say  (x,  y).     In 

B 


B' 
Fig.  38. 

the  right  triangle  F'PD, 

=  PD2  +  FD"2  . 


(i) 


but  PD2  =  f  =         (a2  -  *2)  [from  (A,)], 

and  F'D  =  F'O  +  OD  =  ae  +x  . 


IO4  Analytical  Geometry. 

Substituting  these  values  in  (i). 


FF2  =    -  (a2  -  x2)  +  (ae  +  *)2  =  b2  -  ^-  +  a2  e2 
a2  a 

+  2aex  +  x2  =  tf-  ^  +  a2  -  ^  +  2  aex  +  x\ 
a2 

[since  ?  _    2Ll^]  =  a>  +  a  a**  +  (">-/>*' 
[adding ^-  and  x2]  =  a2  +  2  ae#  +  e2  ^2 

[for    ^^*2=*2*2]. 

/.  F'P=  a  +  ex (i) 

By  similar  process  in  the  right  triangle  FPD, 

FP=a-«* (2) 

Adding  (i)  and  (2).     FT  +  FP  =  2  a. 
Since  F'P  and  FP  are  any  two  focal  radii,  the  sum  of 
the  focal  radii  of  any  point  equals  2  a. 

To  Construct  the  Ellipse. 

ART.  76.  The  definition  of  the  ellipse,  as  a  curve  the 
sum  of  the  distances  of  whose  points  is  constant  and  always 
equal  to  the  major  axis,  gives  us  the  method  of  construction. 

First  Method :  Take  a  cord  the  length  of  the  major  axis, 
and  attach  its  extremities  at  the  two  foci  with  a  pencil 
caught  in  the  loop  thus  formed,  and  keeping  the  cord 
stretched,  describe  a  curve.  It  will  be  an  ellipse,  for  the 
sum  of  the  distances  of  the  pencil  point  from  the  two  points 
of  attachment  (the  foci)  will  always  equal  the  length  of 
the  cord,  that  is,  the  major  axis. 

Second  Method:  Taking  one  of  the  foci  as  centre  and  any 
radius  less  than  the  major  axis,  describe  two  arcs  above 
and  below  the  major  axis,  then  with  the  other  focus  as 


Analytical  Geometry.  105 

centre  and  a  radius  equal  to  the  difference  between  the 
major  axis  and  the  first  radius,  describe  intersecting  arcs. 
These  points  of  intersection  will  be  points  on  the  ellipse,  for 
the  sums  of  their  distances  from  the  foci  will  equal  the 
sum  of  the  radii,  that  is,  the  major  axis.  As  many  points 
as  desired  may  be  located  in  this  way,  and  the  curve  joining 
them  will  be  an  ellipse. 


Fig.  39- 

As  in  Fig.  39  let  A  A'  be  the  major  axis,  F  and  F'  the 
foci.  Taking,  say,  AB  as  radius  and  F'  as  centre  describe 
arcs  m  and  m' . 

Then  taking  A'B  as  radius,  and  F  as  centre  describe 
arcs  n  and  n'\  their  intersections  R  and  S  will  be  points  on 
the  ellipse. 

Taking  any  desired  number  of  points  as  C,  D,  etc., 
perform  the  same  operation,  thus  determining  any  desired 
number  of  points.  A  smooth  curve  through  these  points 
will  be  an  approximate  ellipse. 

ART.  76#.  The  two  following  methods  of  ellipse  con- 
struction are  used  by  draftsmen.  The  first  based  upon 
the  relation  between  the  ordinates  of  points  on  the  ellipse 
and  those  on  the  auxiliary  circles  as  shown  in  Art.  97 
give  a  true  ellipse;  the  second  gives  what  is  known  as  a 
circular-arc-ellipse  and  is  only  an  approximation. 


io6 


Analytical  Geometry. 


First  Method:  Let  O  be  the  centre  of  the  ellipse-  A  A' 
the  major  axis;  BB'  the  minor  axis;  BCB'  the  minor  circle 
and  ADA'  the  major  circle.  (Fig.  390.)  Take  any  num- 
ber of  points  on  the  major  circle  as  R,  S,  T,  etc. 

From  these  points  draw  radii  and  ordinates,  and  through 
the  points  of  intersection  of  the  radii  with  the  minor  circle, 
draw  lines  ||  to  the  major  axis,  AA'.  Where  these  parallels 


D 
Fig.  3Qa. 

intersect  the  ordinates  will  be  points  on  the  ellipse.  The 
points  may  be  made  as  close  together  as  desired  by  draw- 
ing a  great  number  of  radii.  A  smooth  curve  joining  these 
points  will  form  the  ellipse.  Take  the  point  S,  its  radius,  OS, 
and  its  intersection  with  BCB',  P.  Draw  PN. 

In  the  triangle  OSN' 

OP  :  OS  :  :  N'N  :  SN', 

that  is,  b  :  a  :  :  y  :  /,  hence  N  is  a  point  on  the 
ellipse. 

Second  Method:  This  is  known  as  the  three  centre 
method,  or  three  point  method,  and  is  approximate  only. 
Let  AA'  and  BB'  be  the  axes,  intersecting  at  O  (Fig.  396). 


Analytical  Geometry. 


107 


Complete  the  rectangle  BOA'D  and  draw  the  diagonal  A'B. 
From  D  draw  the  line  DE  perpendicular  to  A'B  and  pro- 
duce it  to  meet  BB'  at  C;  with  C  as  a  centre  and  BC  as 
radius  describe  arc  MN;  with  E  (whose  DC  cuts  AA')  as 
centre  and  A'E  as  radius  describe  arc  A'N'. 

With  O  as  centre  and   OB  as  radius  describe  arc   BF, 


cutting  AA'  at  F.  On  A'F  as  diameter  construct  the 
semicircumference  A'B"F,  cutting  B'B  produced  upward 
at  B."  Lay  off  BB"  from  O  toward  B'  to  C'.  With  C  as 
centre  and  CC'  as  radius  describe  arc  RS. 

Lay  OB"  from  A'  on  AA'  to  R'.  With  E  as  centre  and 
ER'  as  radius  draw  arc  R'S',  intersecting  arc  RS  at  T. 
With  T  as  a  centre  and  suitable  radius,  an  arc  described 
will  touch  A'N'  and  MN,  and  complete  the  elliptic  quadrant 
A'B.  A  similar  construction  to  the  right  of  BB'  and  also 
below  AA'  will  complete  the  ellipse. 


io8  Analytical  Geometry. 

EXERCISE. 

What  are  the  axes  and  eccentricities  of  the  following 
ellipses: 

1.  9  x2  +  16  y2  —  144.         3.     x2  +  9  y2  =  81. 

2.  2  *2   +      4  /  =    16.  4.  i  *2  +  $  /  =    I. 

5.  In  an  ellipse,  half  the  sum  of  the  focal  distances  of 
any  point  is  4',  and  half  the  distance  between  foci  is  3'. 
What  is  the  ellipse  equation  ? 

6.  In  a  given  ellipse  the  sum  of  the  focal  radii  of  any 
point  is  10",  and  the  difference  of  the  squares  of  half  this 
sum  and  of  half  the  distance  between  the  foci  is  16.     What 
is  the  equation  to  the  ellipse  ? 

7.  The  eccentricity  of  an  ellipse  is  -f  and  the  distance  of 
the  point  whose  abscissa  is  f  from  the  nearer  focus  is  3. 
What  is  the  equation  to  the  ellipse  ? 

8.  The  major  axis  of  an  ellipse  is  34",  and  the  distance 
between  foci  is  16".     What  is  its  equation  ? 

9.  Find  equation  of  the  ellipse,  in  which    the    major 
axis  is  14"  and  the  distance  between  foci  =  \/3  times  the 
minor  axis. 

10.  In  the  ellipse  2  x2  +  y  2  =  8,  what  are  the  co-ordi- 
nates of  the  point,  whose  abscissa  is  twice  its  ordinate? 
What  are  the  axes? 

11.  What     are     the     co-ordinates    of     the    point,    on 
the  ellipse  4  x2  +  16  y2  —  64,  whose  ordinate  is  3  times  its 
abscissa  ? 

12.  Find   the  intersection   points   of   9  x2  +  16  y2  =  25 
and  2  y  —  x  =  3. 

13.  Find      the     intersection      points      of      the      ellipse 
i6  y2  +  9  ^2  =  288,  and  the  circle  x2  +  y2  =  25. 

14.  In  Ex.  13,  find  the  equation  of  the  common  chord. 

15.  Find  the  angle  between  the  tangents  to  the  ellipse 


Analytical  Geometry. 


109 


and  circle  of  Ex.   13  at  the  point  of  intersection  whose 
co-ordinates  are  both  positive. 

1 6.  An  arch  is  an  arc  of  the  ellipse  whose  major  axis  is 
30',  and  its  chord,  which  is  parallel  to  the  major  axis  and  is 
bisected  by  the  minor  axis,  is  24'  long.     The  greatest  height 
of  the  arc  is  8'.     Find  the  equation  of  the  ellipse  and  plot 
the  arc. 

17.  A  section  of  the  earth  through  the  poles  is  approx- 
imately an  ellipse;  a  section  parallel  to  the  equator  is  a 
circle.     What  is  the  circumference  of  the  Tropic  of  Cancer, 
the  angle  at  the  centre  of  the  earth  between  a  line  to  any 
point  on  it  and  a  line  to  a  point  on  the  equator  being  2^-2^? 

1 8.  If  two  points  on  a  straight  line,  distant  respectively 
a  and  b,  from  its  extremity,  be  kept  on  the  Y-axis  and  X- 
axis,  respectively,  as  the  line  is  moved  around,  the  extremity 
will  describe  an  ellipse,  whose  axes  are  2  a  and  2  b. 

From  this,  suggest  a  method  of  construction  for  the  ellipse. 

ART.  77.     Tangent  to  the  Ellipse. 

The  method  of  finding  the  tangent  equation  is  exactly 


similar  to  that  for  the  circle  and  for  the  parabola.     Taking 
equation  (B) 


no  A  nalytical  Geometry. 

Let  the  points  (V,  /),  (of,  /)  be  on  the  ellipse,  ABA'B', 

say  m  and  n,  then  they  must  satisfy  the  equation 

a2  y2  +  £2  X2      =  a2  pm 

That  is,  a2y'2  +  b2x'2     =  a2b2  .....     (i) 

and  a2/'2+  b2  x"2    =  a2  b2      ....     (2) 

Subtracting  (2)  from  (i); 

a2    (y/2   _    ^2)    +   £2    ^/2 

Factoring  and  transposing, 


whence 


y/  _  y 
Substituting  this  value  of  -—  -  --  ^  in  (B); 


which  is  the  equation  of  the  secant  mw  (Fig.  40).  If 
now  the  point  n  (x",  y")  is  made  to  approach  m  (V,  /), 
when  coincidence  takes  place,  mn  becomes  the  tangent  SR, 
and  (4)  becomes  the  equation  of  the  tangent,  namely, 

b2  x'  ,•         ,. 

y~y  -  ~  ^7(*~"°' 

or  a2  yy'  -  a2  y'2  =    -  b2  xx'  +  b2  x'2. 

a2  yy'+  b2  xx'  =  a2  y'2  +  b2  x'2  =  a2  b2  [by  (i)]  .     (Te) 
Cor.  Letting  y  =  o    in    (T€)    we    get    the    ^-intercept, 

[OM,  Fig.  41!    ' 

The  subtangent,        RM  =  OM  -  OR  =  OM  -  *'.* 
Letting  y  =  o  in  (Te) 

a2  yy'  +  b2  xx'  =  a2  b2, 

x=  —  =  OM. 

x' 

*  It  is  to  be  observed  that  only  length  is  considered  in  estimating 
the  subtangent  and  subnormal,  hence  it  is  unnecessary  to  regard 
the  sign  of  oS. 


Analytical  Geometry. 

2 

Then  subtangent  =  RM  =  — 7  -  x* 


in 


x 

a2  -  x'2 


b2x' 


ART.  78.     Equation  of  the  normal. 
Since  the  normal  is  perpendicular  to  the  tangent  its  slope 
is  the  negative  reciprocal  of  the  tangent  slope,  by  the  rela- 


Fig.  41. 

b2  x' 
The  tangent  slope  is  -    —  —  - 


hence  the  normal  slope  is      -      and  its  equation  will  be 


.    .    .    .    (N.) 


Cor.  Letting    y  =  o    in    (Ne)    we    get    the    ^-intercept 
of  the  normal,  ON,  and  the  subnormal, 

RN  =  OR  -  ON  =  scf  -  ON. 


ii2  Analytical  Geometry. 

Letting     y  =  o  in  (Ne),    y  -  yr  =  (x  - 


-  J2  xf  =  a2  x  -  a2  xf, 


/*2  _7,2  7,2   -/ 

Then         RN  =  x?  -   -  --  -  x'  =  —  . 
a2  a2 

ART.  79.     Slope  equation  of  tangent. 

Let  y  =  mx  -{-  c  .......     (i) 

be  a  secant  line  to  the  ellipse  a2  y2  +  b2  x2  =  a2  b2  (2  ) 

Combining  (i)  and  (2)  to  find  points  of  intersection, 

a2  (mx  +  c)  2  +  b2  x2  =  a2  b2. 
a2  m2  x2  +  2  a2  mcx  +  a2  c2  +  £2  x2  =  a2  b2. 
x2  (a2  m2  +  b2)  +  2  a2  we*  +  (a2  c2  -  a2  b2)  =  o. 

Now  if  this  secant  becomes  a  tangent  the  two  points  of 
intersection,  whose  abscissas  are  .given  by  this  equation, 
become  one  point,  the  point  of  tangency.  As  we  know 
the  condition  that  this  equation  should  have  equal  roots  is 

(a2  m2  +  b2)   (a2  c2  -  a2  b2)  =  (a2  me}2, 
or,     ^<c2  -  a4  m2  b2  +  a2  b2  c2  -  a2  b*  =  ^m2^ 
or  c2  =  a2  m2  +  b2, 

c  =  ±  ^/a2  m2  +  b\ 

Substituting  this  value  of  c  in  (i)  it  becomes  the  equa- 
tion of  the  tangent  in  terms  of  m,  a  and  b,  that  is,  the  slope 
equation  of  the  tangent, 


y  =  mx  ±  ^a2  m2  +  b2       (Te,  m) 

ART.  80.     To  draw  a  tangent  to  the  ellipse. 

It  will  be  observed  that  the  tangent  to  the  ellipse  has  the 


Analytical  Geometry.  113 

same  ^-intercept  as  the  tangent  to  a  circle  having  the 
major  axis  for  a  diameter;  hence  to  draw  a  tangent  to  an 
ellipse  on  the  major  axis  as  a  diameter,  construct  a  circle 
and  produce  the  ordinate  of  the  point  of  tangency  to  meet 
the  circle.  This  point  on  the  circle  and  the  point  of  tan- 


Fig.  42. 

gency  on  the  ellipse  will  have  the  same  abscissa,  and  hence 
the  ^-intercept  of  the  tangents  to  the  circle  at  this  point 
and  to  the  ellipse  will  cut  the  X-axis  in  the  same  point. 

Draw  a  tangent  to  the  circle  at  this  point  and  join  the 
point  of  intersection  with  X-axis  with  the  point  of  tangency 
on  the  ellipse.  The  last  line  will  be  a  tangent  to  the  ellipse 
at  the  required  point.  (Fig.  42.) 

P  =  point  of  tangency;  P'  =  the  point  in  which  the 
ordinate  of  P  cuts  the  circle;  R  =  intersection  of  circle- 
tangent,  RP',  with  the  axis. 

Then  RP  is  the  tangent  to  the  ellipse. 

Supplemental  Chords. 

ART.  81.  The  chords  drawn  from  any  point  on  an 
ellipse  to  the  extremities  of  the  major  axis  are  called  sup- 
plemental chords. 


H4  Analytical  Geometry . 

Let  AP  and  A'P  be  supplemental  chords  of  the  ellipse 
ABA'B'  for  the  point  P.     (Fig.  43.) 

The  equation  of  AP  through  the  point  A  [whose  co- 
ordinates are  (a,  o)],  and  having  say  the  slope  m,  is  [by  (C)] 

y=  m  (x-  a) (i) 

B 


The  equation  of   A'P,  through  the  point  A'  [whose  co- 
ordinates are  (o,  —  a)],  and  having  slope  m',  is  [by  (C)] 

y  =  m'  (x  +  a)  .     .     .     .     .     .     (2) 

multiplying  (i)  and  (2)  together, 

y>  =  mmf  (y*  -a2)      ....     (3) 

which  expresses  the  relation  between  the  co-ordinates  of 
P,  their  intersection.     But  P  (x}  y)  is  on  the  ellipse,  hence 


a2  y2  +  b2x2  =  a2  b2, 


or 


y2  =       (a 


(4) 


Since  (3)  and  (4)  express  the  relation  between  the  co- 
ordinates of  the  same  point,  they  must  be  the  same  equa- 

tion;   hence  comparing;    mm'  =  -    —  ,  which    gives    the 

relation  between  the  slopes  of  supplemental  chords. 

ART.  82.     The  equation  to  a  diameter  of  the  ellipse. 

The  diameter  it  will  be  remembered,  is  the  locus  of  the 
middle  points  of  a  system  of  parallel  chords. 


Analytical  Geometry. 


Let  RS  be  any  one  of  a  system  of  parallel  chords  of  the 
ellipse  ABA'B'  (Fig.  44),  and  T  its  middle  point. 

Let  y  =  mx  +  c  (i)  be  the  equation  of  RS,  and  a2  y2 
+  b2  x2  =  a2  b2  (2)  be  the  ellipse  equation.  Combining  (i) 


and  (2),  we  get  an  equation  whose  roots  are  the  abscissas 
of  R  and  S,  respectively,  if  y  be  eliminated;  an  equation 
whose  roots  are  the  ordinates  of  R  and  S,  if  x  be  eliminated. 


Eliminating  y;        a 


e)2  +  b2  x2  =  a2  b2, 


a2  m2  x2  +  2  a2  mxc  +  a2  c2  +  b2  x2  =  a2  b2, 
2  a2  me      ..  ,    _  2  ^  _  ^2  &2=  Q 


-\ 


x  +  a 


a2  m2  +  b2 

Let  the  two  roots  of  (3)  be  represented  by  x'  and  x?. 
Then  by  the  structure  of  a  quadratic, 

2  a2  me 


(3) 


b2 


Calling  the  ordinates  of  T,  (X,  Y), 


then        X 


x' 


a  me 


2  a2  m2  +  b2 

Eliminating  x  from  (i)  and  (2) 


(4)  [by  Art.  32] 


a2y2 


a2  y2  +  Z>2  ^ 
b2y2  -  2  b2  yc 


m 


=  a2  b2, 
=  a2  b2, 


1  1  6  A  nalytical  Geometry. 

a2  m2  y2  +  b2  y2  -  2  b2  yc  +  b2  c2  =  a2  b2  m2, 


Calling  the  two  roots  of  (5),  yf  and 

2b2C 


=  + 


a2  >«2  +  b2 

and  Y  =  ^m-  =    ,+.y.CM     -•-.     (6) 

2  a2  w2  +  62 

Since  c  is  a  variable  it  must  be  eliminated  between  (4) 
and  (6),  for  we  must  express  the  relations  between  the 
co-ordinates  of  these  mid-points  of  the  chords  in  terms  of 
constants  to  get  the  true  equation  of  their  locus. 

Divide  (6  by  (4) 


Y  _  a2  m2  +  b2  __  -  b2 
X          -  a2  me         a2  m 

a2  m2  +  b2 


is  the  equation  of  the  diameter,  since  it  expresses  a  constant 
relation  between  the  co-ordinates  of  the  mid-point  of  RS, 
and  RS  stands  for  any  one  of  the  parallel  chords,  m  is  a 
constant  because  the  chords  being  parallel,  all  have  the 
same  slope.  The  form  of  this  equation  shows  that  the 
diameters  pass  through  the  centre,  since  the  constant  or 
intercept  term  is  missing. 

Since  this  equation  represents  any  diameter  whatever, 
it  follows  that  any  chord  passing  through  the  centre  of  the 
ellipse  is  a  diameter,  and  hence  bisects  a  system  of  parallel 
chords. 


Analytical  Geometry.  117 

Conjugate  Diameters. 
ART.  83.     It  will  be  observed  in  the  equation 

y  =  —  — —  x,  the  slope  is  —  — — ;  that  is,  it  is 

a2  m  a2  m  a2 

divided  by  m,  the  slope  of  the  chords. 

If  a  system  of  chords  be  drawn  parallel  to  this  first  diam- 
eter, their  slope  will  be  that  of  this  diameter,  namely, 

b2 

a2  m 

The  slope  of  the  diameter  corresponding  to  this  system 
of  chords,  by  above  principle,  will  be 

b2  b2 

1   *    ~    ~T~  =  m- 

a2  a2m 

Hence  the  equation  of  this  second  diameter  is  y  =  mx. 

The  slope  of  this  diameter  is  the  same  as  that  of  the 
chords  of  the  first;  hence  each  is  parallel  to  the  chords  of 
the  system  determining  the  other. 

Such  diameters  are  called  conjugate  diameters  and  are 
determined  by  the  condition  that  the  product  of  their 
slopes  is, 


ART.  84.    Tangents  at  the  extremities  of  conjugate  diameters. 

The  farther  a  chord  is  from  the  centre  the  nearer  together 
are  its  intersection  points  with  the  ellipse,  evidently.  Since 
the  mid-point  must  always  lie  between  these  intersection 
points,  in  any  system  of  parallel  chords,  as  the  chords  are 
drawn  farther  and  farther  from  the  centre,  their  points  of 
intersection  and  their  mid-points  approach  coincidence, 
and  eventually  the  chord  becomes  a  tangent  at  the  end  of 
the  diameter,  when  the  three  points  coincide. 


n8 


Analytical  Geometry. 


Hence  the  tangent  at  the  extremity  of  a  diameter  is 
parallel  to  its  system  of  chords.* 

This  fact,  combined  with  the  relation  between  conjugate 
diameters,  defined  in  Art.  83,  enables  us  to  readily  draw 
any  pair  of  conjugate  diameters.  Thus:  at  the  extremity 
of  any  diameter  draw  a  tangent  to  the  ellipse;  the  diameter 
drawn  parallel  to  this  tangent  will  be  the  conjugate  to  the 
given  diameter. 

ART.  85.  The  co-ordinates  of  extremities  of  a  diameter 
in  terms  of  the  co-ordinates  oj  the  extremity  of  its  conjugate. 


Fig.  45. 


Let  the  co-ordinates  of  R,  the  extremity  of  the  diameter 
RS,  be  (V,  /),  to  find  the  co-ordinates  of  R'. 


*  This  may  be  shown  analytically  thus:     The  intersection  point 

b2 
of  the  diameter  y  =    —  — = —  x   with   the   ellipse  a2  y2  +  b2  x2 


a2  b2,  is  (by  combining  equations)  x' 
b2 


and    y' 


\/a2  m2  +  b2 
Taking  the  tangent  equation  (T«),  and  substituting 


these  points  for  points  of  tangency,  we  find  the  slope  of  the  tangent 
at  x',  y't  to  be  m,  but  this  is  the  slope  of  the  chords.  Hence  tangent 
is  parallel  to  chords. 


Analytical   Geometry.  119 

Draw  the  tangent  (Fig.  45)  MN  at  R.  By  (T.)  its 
equation  is  a2  yyf  +  b2  xx'  =  a2  b2. 

Then  the  equation  to  R'S'  is  a2  yyf  +  b2  xxf  =  o  .  .  (i) 
since  it  is  parallel  to  MN,  but  is  drawn  through  the  origin, 
hence  the  absolute  term  is  o. 

Let  the  ellipse  equation  be  as  usual,  a2^  +  b2  x2  =  a2  b2. 

Since  (#',/)  is  on  the  ellipse; 

a2  y'2  +  b2  x'2  =  a2  b2     .....     (2) 

If  (i)  and  the  ellipse  equation  be  combined,  the  resulting 
values  of  x  and  y  will  be  the  co-ordinates  of  the  points  of 
intersection,  R'  and  S'. 

Substituting  the  value  of  y  from  (i)  in  the  ellipse  equation, 


a2  y2 

b2  x2  (b2  x'2  +  a2  y'2}    .  .  a,  p 
a2/2 

b2x2(a2b2}    __a,b, 
a2/2 

[Since  b2  x'2  +  a2  y'2  =  a2  b2, 
point  (xf,  /)  being  on  the  ellipse.] 


Whence 

b2 

.  a/ 
*-±f> 

and  hence  y  =  -F  — 


120 


Analytical  Geometry. 


ART.  86.     The  length  of  conjugate  diameters.    Draw  the 
co-ordinates  RT  and  R'T'  of  R  and  R'  respectively,  R  and 

B 


R'  being  the  extremities  of  conjugate  diameters.  (Fig.  46.) 
Then  if  (OT,  RT)  are  (-  *',  /),  (OT',  R'T')  are 


In  the  right  triangles   ORT  and   OR'T' 
OR2  = 

,2 


and  OR'2  =  OT'2  +  R'T'2  = 
Then          OR2  +  OR'2  = 


+ 


5r  +  /'  + 


b2 


_  b2  x'2  +  a2  y'2       a*  y'*  +  b2  x'2 

b2  a2 

a2  b2      a2  b2       .      „ 

^r  +  -^=-2  +  ^ 

for  since  (xr,  /)  is  on  the  ellipse, 

b2  x'2  +  a2  y'2  =  a2  b2. 

That  is,  the  sum  of  the  squares  of  any  pair  of  conjugate 
diameters  equals  the  sum  of  the  squares  of  the  axes. 


Analytical  Geometry. 


121 


Conjugate  diameters  are  usually  represented  by  a'  and 
&',  hence 

a'2  +  b'2  =  a2  +  b2. 

ART.  87.     Major  and  Minor  auxiliary  circles. 

The  circle  drawn  with  the  major  axis  as  diameter  is 
called  the  major  auxiliary  circle. 

The  circle  drawn  with  the  minor  axis  as  diameter  is  called 
the  minor  auxiliary  circle. 

Fig.  47,  the  angle  AOP',  is  called  the  eccentric  angle  of 
the  point  P  on  the  ellipse. 

The  eccentric  angle  of  any  point  is  determined,  thus: 

Produce  the  ordinate  of  the  given  point   to  meet  the 


Fig.  47. 

major  auxiliary  circle,  and  join  this  point  of  meeting  on 
the  circle  with  the  centre.  The  angle  between  this  joining 
line  and  the  axis,  measured  positively,  is  the  eccentric  angle 
of  the  point  on  the  ellipse. 

ART.  88.     Relation  between  the  ordinates  of  a  point  on 
the  ellipse  and  of  the  corresponding  point  on  the  major  circle. 

The  equation  of  the  major  circle,  whose  radius  is  a,  is, 

x2  -j-  y2  =  a2  or  y2  =  a2  —  x2      .     .     .     (i) 


122  Analytical  Geometry. 

Call  the  Point  P'  (Fig.  47),  (V,  /')  and  P,  (*',  /). 

(Observe  P'  and  P  have  the  same  abscissa.) 

Then  from  (i),  y"2  =  a2  —  x'2 (2) 

b2 
Also,    y'2  =  -    (a2  —  x'2)   (3)   (from  ellipse  equation). 

Dividing  (3)  by  (2) 

y^_        P 

f2   ==  a2' 

or  2_  =  — ,     whence  yf  :  y"  :  :  b  :  a. 

y"      a 

That  is,  the  ordinate  oj  any  point  on  the  ellipse  is  to  the 
ordinate  of  the  corresponding  point  on  the  major  circle  as 
the  semi-minor  axis  is  to  the  semi-major  axis. 

Corollary:  Let  Q  be  the  intersection  of  OP'  with  the  minor 
circle.  (Fig.  47.) 

Join  Q  with  P. 

Then  since  OQ  =  b  and  OP'  =  a, 
and          /  :  f  :  :  b  :  a,  /  :  /'  :  :  OQ  :  OP', 

or  PD  :  P'D  :  :  OQ  :  OP'. 

That  is,  QP  is  parallel  to  OD;  that  is,  parallel  to  the 
axis. 

Hence  RP,  the  prolongation  of  QP,  to  BB',  equals  OD  = 
the  abscissa  of  P  and  P'.  This  furnishes  another  method 
of  drawing  an  ellipse.  Thus: 

Draw  two  concentric  circles  with  the  given  major  and 
minor  axes  as  diameters,  respectively,  in  their  normal 
positions. 

Make  any  angle  with  the  major  axis,  as  AOP'  in  Fig.  47, 
and  let  the  terminal  line  of  this  angle  intersect  the  two 
circles  in  Q  and  P'  respectively.  Then  the  intersection  of 
the  abscissa,  RQ,  of  Q,  with  the  ordinate,  P'D,  of  P',  will 
be  a  point  on  the  ellipse. 


Analytical  Geometry. 


123 


This  may  be  shown  by  analytical  means,  purely,  for 
(Fig.  47)  in  the  right  triangle  OP'D,  OD  (=  RP)  = 
OP'  cos  P'OD  =  a  cos  <£,  say,  and  drawing  QE  perpen- 
dicular to  OA, 

PD  =  QE  =  OQ  sin  QOD  =  b  sin  0,  but  the  values 
a  cos  <f>  for  xt  and  b  sin  <£  for  y,  satisfy  the  ellipse  equation. 

a2  y2  +b2x2=  a2  b2, 

thus,  a2  b2  sin2  0  +  a2  b2  cos2  <f>  =  a2  b2, 

sin2  (f>  +  cos2  <f>=  i, 

hence  since  OD  and  PD  are  the  co-ordinates  of  P,  P  is  on 
the  ellipse. 

ART.  89.  The  eccentric  angle  between  two  conjugate 
diameters. 

Let  the  eccentric  angle  of  R'   (yf,  /),  the  extremity  of 

R'S'  be  0,  and  that  of  R  /  -  -^ ,  +  —  V  the  extremity  of 


Fig.  48. 

the  conjugate  diameter  RS  be  <f>.     (Fig.  48.) 
Then  in  the  right  triangle  OP'T', 


124  Analytical  Geometry . 


cos  P'OT'  =   p£~    or  cos  d  =  -  .     .     (i) 

In  the  right  triangle  OPT, 

sin   P"OT   =  ^.  =  *__-.     .     [Art.  88] 


That  is,  sin  (180  —  <j>)  =  sin  6  =       x     -  =  —    .  .   (2) 

a  a 

.'.  sin  (£  =  cos  6  from  (i)  and  (2), 
whence  by  trigonometry, 

(j)  =  90  +  0    or    <j)  —  6  =  90°. 

That  is,  the  difference  between  the  eccentric  angles  of 
the  extremities  of  conjugate  diameters  is  a  right  angle. 

ART.  90.  By  combining  the  slope  equations  of  two 
perpendicular  diameters,  both  expressed  in  terms  of  the 
slope  of  one,  it  is  readily  proved,  as  was  done  under  the 
parabola,  that  the  locus  oj  their  'intersections  is  a  circle, 
whose  equation  is 

x>  +  y*  =  a2  +  b2. 

This  circle  is  called  the  director  circle.  Also  by  a  similar 
process  it  can  be  shown  that  the  major  auxiliary  circle  is 
the  locus  of  the  intersection  oj  a  tangent  with  the  perpendic- 
ular to  it  from  a  focus. 

ART.  91.  The  ellipse  possesses  a  physical  property, 
somewhat  similar  to  that  possessed  by  the  parabola,  namely: 

The  angle  formed  by  the  focal  radii  to  any  point  on  the 
ellipse  is  bisected  by  the  normal  at  that  point. 

Geometry  tells  us  that  the  bisector  of  an  angle  of  a  tri- 
angle divides  the  opposite  side  into  segments  proportional 


Analytical  Geometry. 


125 


to  the  other  sides,  hence,  if  we  can  prove  (Fig.  49)  that 
F'N  :  FN  :  :  F'P  :  FP  our  proposition  is  established.  It  is 
necessary  then  to  find  values  for  these  four  lines  in  the 
same  terms.  ON  the  ^-intercept  of  the  normal  was  found 
in  Art.  78,  Cor.  to  be 


a2-  b2 


xf  =  e2 


where  xf  is  the  point  of  tangency. 


Fig.  49. 

Let  P  (Fig.  49)  be  (*',  /). 

Then  F'N  =  F'O  +  ON  =  c  +  cV  =  ae  + 


(since  —  =  e,  hence  c  =  ae\ 
a 

FN  =  FO  -  ON  =  ae  -  e*x'. 

F'P  =  a  +  ex'  and  FP  =  a  -  ex'      .     . 

ae  -f  e2x'   _  e  (a  +  e'x')  _  a  +  exf 


(Art.  75) 


But 


ae  —  e2xf       e  (a  —  exf)      a  —  exf 
F'N         F'P 


FN 


FP 


or    F'N  :  FN  :  :  F'P  :  FP. 


It  follows  from  the  law  of  reflection  for  vibrations,  that 
if  light  or  sound  issue  from  one  focus  of  an  ellipse  it  will 
be  reflected  to  the  other  focus. 


126 


Analytical  Geometry. 


ART.  92.     The  area  oj  an  ellipse. 

Draw  the  major  auxiliary  circle  to  the  ellipse  ABA'B', 
and  construct  rectangles  as  indicated  in  Fig.  50.  . 

Then  the  area  of  one  of  these  rectangles  in  the  ellipse  as 
mnpo  is 

Area  mnpo  =  mn  X  pn. 

Let  the  points  on  the  ellipse  beginning  with  p  be  (V,  /), 
(V,  /'),  (X",  /"),  etc.,  and  the  corresponding  points  on 
the  circle  beginning  with  R,  be  (xf,  y^,  (x",  y2),  (x"f,  y3)  etc. 

Then  Area  mnpo  =  (V  —  x")  yf. 

The  corresponding  rectangle  in  the  circle 

mnRS  =  (xr  -  x")  yv 

.    mnRS  =  Ix'  -  x"\  y_v  =  ^  =  a 
mnpo       \xf  —  x"J  yf       y'      b' 

As  this  is  a  typical  rectangle  each  circle  rectangle  is  to 
each  ellipse  rectangle  as  a  is  to  b,  hence  by  the  law  of  con- 
tinued proportion,  the  sum  of  all  the  circle  rectangles  is  to 
the  sum  of  all  the  ellipse  rectangles  as  a  is  to  b. 

As  the  above  expression  is  independent  of  the  size  or 


Fig.  50. 

number  of  the  individual   rectangles   the  relation  is  the 
same  when  the  number  of  rectangles  becomes  infinite.     But 


Analytical  Geometry.  127 

in  this  latter  case  the  sum  of  the  areas  approach,  respec- 
tively, the  area  of  the  circle  and  that  of  the  ellipse;  hence, 
finally, 

Area  of  the  circle  _  a 
Area  of  the  ellipse       b 

That  is,  area  of  the  ellipse  =  —  times   the   area  of   the 

a 

circle,  but  area  of  the  circle  =  ?ra2. 

.'.  area  of  the  ellipse  =    •  -   .    no2  =  nab. 


ellipses 


EXERCISE. 

What  are  the  equations  of  the  tangents  to  the  following 
ipses  ? 

1.  x2  +  4  y2  =  4  at  the  point   (f  ,  i). 

2.  4  x2  +  9  y2  =  36  at  the  point  (i,  f  \/2). 

3.  x*  +  3  /  =  3  at  the  point   (f  ,  i). 

4.  9  x2  +  25  y2  =    225  at  the  point  (4,  ?). 

5.  25  x2  +  100  y2  =  25  at  the  point    (?,  2). 

6.  x2  +  2  y2  =  18  at  the  point  (?,  i). 

7.  Find  the  normal  equation  to  the  above  ellipses. 

8.  What  are  the  equations  of  the  tangents  to  the  ellipse 
16  y2  +  9  x2  =  144  from  the  point  (—3,  2)  ? 

9.  What  is  the  equation  of  the  tangent  to  the  ellipse 
9  x2  +  25  y2  =  225,  that  is  parallel  to  the  line  10  y—  8  x  =  5. 

10.  What  is  the  equation  of  the  tangent  to  the  ellipse 

x2  +  4  y2  =  4,  that  is  parallel  to  the  line  *-  —  x  \/3  =  i  ? 

2 

11.  What  is  the  equation  of  the  tangent  to  the  ellipse 
4  x2  +  9  y*  =  36,    which    is    perpendicular    to    the    line 
^-  3  *=  5? 


128  Analytical  Geometry. 

12.  The  subtangent  to  an  ellipse,  whose  eccentricity  is 
§,  is  |.     What  is  the  ellipse  equation? 

13.  Find  the  equation  of  the  tangent  to  the  ellipse  in 
terms  of  the  eccentric  angle  of  the  point  of  tangency. 

14.  What  are  the  equations  of  the  tangents  to  the  ellipse 

x2      y2 

—•  +  —  =  i,  which  form  an  equilateral  triangle  with  the 

9        4 
axis? 

15.  What  is  the  equation  of  the  diameter  conjugate  to 
4^  +  9^=o? 

16.  2  y  -\-  x  =  12  and  2  y  =  i  #  +  3  are  supplementary 
chords  of  an  ellipse.     What  is  its  equation  ? 

17.  The  middle  point  of  a  chord  of  the  ellipse  25  y2  +  9  x2 
=  225  is  (—  5,  i).     What  is  the  equation  of  the  chord? 

18.  The  equation  of  a  diameter  to  the  ellipse  4  x2  +16  y2 
=  64  is  4  y  =  x.      What  is  the  equation  of  a  tangent  to  the 
ellipse  at  the  end  of  its  conjugate  diameter  ? 

19.  Find   the   equation   of   the   tangents   to   the   ellipse 

O  *V^  *\7 

« — | —  =  i,  which  makes  an  angle  whose  tangent  is  3 
16       9 

with  the  line  2  y  =  x  —  i. 

20.  Find  the  equation  of  the  normal  to  the  ellipse  x2  + 
4  y2  =  4?  which  is  parallel  to  the  line  4  x  —  37=  7. 

21.  Show  that  the  product  of  the  perpendiculars  from 
the  two  foci  upon  any  tangent  is  equal  to  the  semi-minor 
axis. 

22.  Find    the    equation    to    a    diameter    of    the    ellipse 

x2         y2 

h  *-     =    i,  which    bisects    the    chords    parallel    to 

16         9 

3  x  -  5  y  =  9. 

23.  Find  the  locus  of  the  centres  of  circles  which  pass 
through  (i,  3)  and  are  tangent  internally  to  x2  +  y2  =  25. 


Analytical  Geometry.  129 

x2  v2 

24.  The  equation  of  an  ellipse  is  +    —      =     i. 

169         144 

What  is  the  eccentric  angle  of  the  point  whose  abscissa 
is  5? 

25.  Find  the  equation  of  the  chord  joining  the  points 
of  contact    [called  the  chord  of  contact]  of  two  tangents  to 
the  ellipse  9  x2  +  16  y2  =  144,  drawn  from  (4,  3)  outside 
the  ellipse. 

26.  Find  the  locus  of  the  vertices  of  triangles  having  the 
base  2  a,  and  the  product  of  the  tangents  of  their  base 

i       & 
angles    -  . 
c 

27.  The  minor  axis  of  an  ellipse  is  18,  and  its  area  is 
equal  to  that  of  a  circle  whose  diameter  is  24.     What  is 
the  equation  to  the  ellipse  ? 

28.  The  axes  of  an  ellipse  are  40  and  50.     Find  the 
areas  of  the  two  parts  into  which  it  is  divided  by  the  latus 
rectum. 


CHAPTER  VII. 
THE   HYPERBOLA. 

ART.  93.  The  characteristic  of  the  hyperbola  is  that 
the  difference  of  the  distances  of  any  point  on  it,  from 
two  fixed  points,  is  constant. 

With  this  understanding  of  the  locus, 

To  find  the  equation  oj  the  hyperbola. 

In  Fig.  51,  let  P  be  any  point  on  the  hyperbola,  whose 
foci  are  F  and  F',  and  whose  vertices  are  A  and  A'.  Draw 
the  ordinate  PD  and  the  focal  radii  PF,  PF'. 


Fig.  Si. 

The  co-ordinates  of  P  are  (OD,  PD),  say  (x,  y),  O  being 
the  origin,  OX  and  OY  the  axes.     It  is  our  problem  then 


Analytical  Geometry.  131 

to  find  a  relation  between  OD  and  PD,  and  the  right  tri- 
angle PFD  suggests  itself. 

In    the    right    triangle    PFD,    PF2  =  PD2  +  FD2  (i). 
Call  the  focal  distance  OF,  c.     Then  (i)  becomes, 
PF2  =  r2  =  y2  +  (x  _  cy  |-since  FD  =  OD  -  OF  =  x  -c] 

(x-c)2  .....     (2) 


In  the  right  triangle  F'PD, 

PF'2  ==  PD2  +"F7D2.     That    is,  r'2  =  y2  +  (x  +  c)2  [since 
FD  =  OD  +  OF'  =  x  +  c]    or   /  =  \/y2  +  (x+c)2  (3) 

By  definition,     /  —  r  =  constant  =  2  m,  say. 

Subtract  (2)  from  (3); 


-  r  =  2  m. 
Transpose  and  square; 


+^+  2cx  +  ^{=  4m2  +  4  m  \/y2  +  (x  - 

+  ;X+^-  2  ex  +  <H 
Transpose,  collect,  and  divide  by  4; 

m  \/y2  +  (x  ~~  CY  =  ex  —  m2. 
Square  again; 


m2  y2  +  m2  x2  —  2  wS?&.  +  m2  c2  =  c2  x2  —  2 
Collect;  m2  y2  +  (m2  -  c2)  x2  =  m2  (m2-c2)      .     .     (4) 

To  determine  m  it  is  only  necessary  to  give  x  and  y 
suitable  values,  or  rather  to  give  y  the  particular  value  o, 
since  the  above  equation  is  true  for  every  point  on  the 
hyperbola.  We  then  get  the  value  of  x  for  the  vertex,  since 
the  ordinates  of  A  and  A'  are  o. 

Letting  y  =  o  in  (4) 

(m2  -  c2)  x2  =  m2  (m2  -  c2\ 


132  Analytical  Geometry. 

whence  x2  =  m2;  x  =  ±  m, 

but  x  here  equals  O A     or     OA', 

hence  m  =  OA  or  OA'; 

that  is,  2  m  =  the  major  axis  AA'.     As  in  the  ellipse 

call  AA',  2  a;  then  m  =  a,  and  (4)  becomes, 

a2  y2  +  (a2  -  c2)  x2  =  a2  (a2  -  c2)   .     .     .     (5) 
Let   ,  c2  -  a2  =  b2, 

which  by  analogy  with  the  ellipse  we  may  call  the  minor 
axis.     We  shall  see  that  this  is  justified.     Then  (5)  becomes, 

a2  y2  -  b2  x2  =  -  a2  b2, 
or  b2x2-a2y2=  a2b2 (A») 

ART.  94.     A  glance  at  the  figure  will  show  that  c  is 
greater  than  a,  hence  the  eccentricity, 

e  =    -  is  >  i. 
a 

Then  in  the  polar  equation  for  conies 

P  =   e± j  (e  >  i), 

i  -  e  cos  0 

and  by  a  process  exactly  like  that  in  Art.  84,  this  becomes 
for  the  hyperbola, 

a(e2-  i) 


P  = 


i  —  e  cos 


ART.  95.     To  determine  b  in  the  figure  of  a  hyperbola. 

The  relation  c2  —  a2  =  b2,  immediately  suggests  a  right 
triangle  with  c  as  hypotenuse.  Hence  with  c  as  radius 
and  A  or  A'  as  centre,  describe  arcs  cutting  the  y-axis  at 
B  and  B',  OB  will  equal 

b,  or  BB'  =  2  6;  for  OB2  =  AB2  -  OA2  =  c2  -  a2. 


Analytical  Geometry.  133 

It  is  plain  that  the  curve  does  not  cut  this  minor  axis, 
for,  setting  x  =  o  [the  abscissa  of  any  point  on  BB'  =  o] 

in  (A,), 

-  a2  y2  =  a2  b2 


y  =  ±  A/— b2  =  ±  b\/  —  i,  an  imaginary  value. 

ART.  96.     To  find  the  length  of  the  focal  radii  for  any 
point,  r  and  /. 


or 


Fig.  sia. 

In  Fig.  510,  PF2  =  r2  =  PD2  +  FD2, 

r2  =  y2  +  (x  -  c)2   .     . 


Since 


e  =  —  »        c  =  ae, 
a 


and  (i)  becomes, 

r2  =  f  +  (X  -  ae}\ 
or  r2  =  y2  +  x2  —  2  aex  +  a2  e2. 


(i) 


134  Analytical  Geometry. 

By        (Ah),y2  =  ¥   (*2-<z2). 

/.  r2  =  ^  -  62  +  *2  -  2  ae*  +  a2 


(fl2  +  fr2)  *2  -  b2  -  2  aex  +  a2  e2.     [But  a2  +  b2 


=  c2]     =    •  -  b2  —  2  aex  +  a2  e2  =  e2  x2  —  2  tf&v 

<r 


+  a2e2  -  b2  =  e2  x2  -  2  aex  +  a2  [since  a2  e2  -  b2 


.......        (3) 

By  exactly  similar  treatment  of  (3)  Art.  93,  we  get, 

/  =  ex  +  a      .......     (4) 

Subtract  (3)  from  (4),  rf  —  r  =  2  a,  which  shows  that 
the  constant  difference  rf  —  r  is  always  equal  to  the  major 
axis. 

ART.  97.  A  comparison  of  the  ellipse  and  hyperbola 
equations  shows  that  if  in  the  ellipse  equation  —  b2  is  sub- 
stituted for  +  b2,  the  hyperbola  equation  results;  hence 
since  the  fundamental  processes  in  deriving  tangent,  nor- 
mal, and  diameter  equations  are  the  same  for  all  curves, 
the  equations  for  these  lines  in  relation  to  the  hyperbola 
can  be  derived  from  the  corresponding  equations  in  the 
ellipse  by  substituting  —  b2  for  b2. 


Analytical  Geometry.  135 

For  example : 

(a)  The  ellipse  tangent  has  the  equation, 
a2  yy  +  b2  xx>  =  a2  62j 

hence  the  hyberbola  tangent  is, 

a2yyf  -  b2  xxf  =  -  a2b2 

or  b2  xxf  -  a2yyf  =  a2b2  .     .     .     .     (Th) 

The  slope  form  is, 

y  =  mx  ±  ^/a2m2  -  b2  .    .     .     .     (TA  J 
(6)  The  normal  equation  for  the  ellipse  is, 


hence  the  normal  equation  for  the  hyperbola  is, 

_  ,  =  _  a2y  (x  _  x/] 

b2  x' 

(c)  The  subtangent  then  is — - ,  and  the  subnormal 

00 

b2  x' 
is   -     -  ,  the  same  as  for  the  ellipse. 

a2 

(d)  The  equation  for  a  diameter  of  the  ellipse  is, 

v-  —  *, 

a2m 
hence  a  diameter  to  the  hyperbola  is, 

b2 
a2  m 

Conjugate  diameters  are  defined  in  the  same  way,  hence 
the  product  of  their  slopes,  m  and  m',  say,  is 

mm'  =  —  [—  b2  replaces  b2]. 


136  Analytical  Geometry. 

ART.  98.  As  the  ellipse  becomes  a  circle  when  its  axes 
become  equal,  for  when  b  =  a, 

a2  y2  +  b2  x2  =  a2  b2  becomes  y2  +  x2  =  a2, 

so  if  the  axes  of  a  hyperbola  become  equal,  we  call  it  an 
equilateral  hyberbola,  which  is  the  hyperbola-analogue  of 
the  circle. 

In  b2  x2  —  a2  y2  =  a2  b2,  let  b  =  a\  then  x2  -  y2  =  a2 
is  the  equation  of  an  equilateral  hyperbola. 

ART.  99.  The  latus  rectum  of  the  hyperbola  is  readily 
found  from  its  equation  by  setting 


x  =  ±  c  =  ±  \/a2  +  b2. 
Whence  b2(a2  +  b2)  -  a2y2  =  a2b2 


b-      =  +  bl 

a2 '  a 


2  b 

2y=  —  =    latus  rectum,  since  it  is  the 
a 


double  ordinate  through  the  focus. 


EXERCISE. 

What  are  the  axes  and  eccentricities  of  the  following 
hyperbolas : 

i.   2  x2  —  3  y2  =  9.  2.   x2  —  4  y2  =  4. 

3.    16  y2  -  9  x2  =  144.  4.    5  x2  -  8  f  =  15. 

5.    9  y2  -  4  x2  =  -  36.  6.    4  y2  -  3  x2  =  12. 

7.   a;2  —  16  y2  =  16.  8.   4%2  —  i6y2  =  —  64. 

9.  What  is  the  equation  of  a  hyperbola,  if  half  the  dif- 
ference of  the  focal  radii  for  any  point  is  7,  and  half  the 
distance  between  foci  is  9  ? 


Analytical  Geometry.  137 

10.  What  is  the  equation  of  the  hyperbola,  whose  con- 
jugate axis  is  6  and  eccentricity,  ij? 

11.  The  co-ordinates  of  a  certain  point  on  a  hyperbola, 
whose  major  axis  is  20,  are  x  =  6,  y  =  4.     Find  its  equa- 
tion. 

12.  The  eccentricity  of  a  hyperbola  is  if,  and  the  longer 
focal   radius  of  the  point  x  =  5,  is  32.     Find  hyperbola 
equation. 

13.  In  a  hyperbola  2  a  =  20,  and  the  latus  rectum  =  5 
Find  its  equation. 

14.  The  conjugate  axis  =  10,  and  the  transverse  axis  is 
twice  the  conjugate.     Find  the  equation. 

15.  The   conjugate   axis  =  16   and   the   transverse   axis 
=  f  of  the  distance  between  foci.     Find  the  equation. 

16.  In    the     hyperbola     25  x2  —  4  y2  =  100,     find    the 
co-ordinates  of  the  point  whose  ordinate  is  2$  times  its 
abscissa. 

17.  In   the   hyperbola   25  x2  —  169  y2  =  4225,   find  the 
focal  radii  of  the  point  whose  ordinate  is  10  V2- 

Find  the  intersection  points  of  the  following  : 

18.  16  y2  —  4  x2  =  16  and  2  x  —  y  =  3. 

19. 4-2L  =  L    and  $y—  2x  +  8=o. 

499 

20.  9  y2  —  16  x2  =  144  and  x2  +  y2  =  36. 

21.  9  y2  —  6  x2  =  36  and  4  x2  +  9  y2  =  36. 

22.  16  x2  —  25  ^2  =  400  and  4X2  -\-  16  y2  =  16. 

23.  #2  —  y2  =  —  50  and  #2  +  y2  =  100. 

24.  Find  the  equation  of  the  tangent  to  the  hyperbola 
16  y2  —  9  x2  =  144  at  the  point  (V5,  5). 

25.  At    what   angle   do   the   curves    in    Ex.    22    inter- 
sect? 


138  Analytical  Geometry. 

CONSTRUCTION    OF    THE    HYPERBOLA. 

ART.  100.  The  definition  of  the  hyperbola  suggests  a 
method  of  mechanical  construction  similar  to  that  for  the 
ellipse. 

Since  the  difference  between  the  focal  radii  is  constant, 
if  a  fixed  length  of  string  be  taken,  attached  at  the  two 
foci,  and  the  same  amount  subtracted  from  each  of  two 
branches,  continually,  the  hyperbola  results. 


Fig.  52' 

In  Fig.  52,  let  a  straight  edge  of  length  /  +  2  a,  be 
pivoted  at  F',  and  one  end  of  a  string  of  length-  /  be  fastened 
to  its  free  end,  N,  and  attached  to  the  focus  F,  at  its  other 
end. 

A  pencil  pressed  against  the  straight  edge,  keeping  the 
string  stretched  (as  at  P),  will  describe  the  right  branch 
of  the  hyperbola.  For  at  any  point  as  at  P, 

PF'  -  PF  =  (F'N  -  PN)  -  (NPF  -  PN)  = 
F'N  -  NPF  =  /  +  2  a  -  I  =  2  a. 


Analytical  Geometry. 


'39 


The  other  branch  may  be  described  similarly  by  pivot- 
ing at  F,  and  attaching  the  string  at  F'. 

Second  Method :  The  hyperbola  may  also  be  constructed 
by  points,  making  use  of  the  definition.  Let  AA'  [Fig.  52 
(a)]  be  the  major  axis,  F  and  F'  the 'foci  and  O  the  centre. 


Fig.  5«a. 


Let  LK  [Fig.  52  (b)]  =  AA'.     Extend  LK  and  take  any 
number  of  points  on  LK  produced  as  P,  R,  S,  T,  etc.     With 


P    R 


Fig.  52b. 

LP  >  LK  as  radius  and  F  and  F',  successively,  as  centres 
describe  arcs  as  at  G,  H,  G'  and  H';  with  the  same  centres 
and  KP  as  radius,  describe  intersecting  arcs  at  G,  H,  G' 
and  H'.  The  intersections  will  be  points  on  the  ellipse  for 
the  radii  LP  -  KP  =  LK  =  AA'.  The  same  process 
with  points  R,  S,  T,  etc.,  will 'give  as  many  points  as  desired. 
A  smooth  curve  through  these  points  will  be  the  hyperbola. 


140  Analytical  Geometry. 

CONJUGATE   HYPERBOLA. 

ART.  TOI.  The  hyperbola  whose  axis  coincides  with 
the  axis  of  ordinates  is  called  the  conjugate  hyberbola  to  the 
one  whose  axis  is  the'^-axis.  MEN  —  RB'S  (Fig.  53). 


Fig.    53- 

Its  equation  is  readily  found  to  be 

ay  -  b2oc2  =  a2b2. 

ART.  102.  If  the  equations  of  two  conjugate  diameters 
be  combined  with  the  equation  to  the  original  hyperbola, 
it  will  be  found  that  the  results  will  be  imaginary  for  one 
of  the  diameters,  showing  that  both  diameters  do  not 
touch  the  original  hyperbola.  Thus: 

Let  y  =  mx (i) 


Analytical  Geometry.  141 

and  .     .       »-£ <•> 

be  conjugate  diameters. 
Combining  these  with 

b2x2  -  a2y2  =  a2b2 (3) 

we  get  from  (i)  and  (3), 

a2b2 


b2  -  a2m2  ' 
from  (2)  and  (3), 


a2m2  -  b2 

If  b2  —  a2m2  is  plus,  a2m2  —  b2  must  be  minus,  hence 
if  the  first  x2  is  plus,  and  hence  x,  real,  the  second  x2  is 
minus,  and  hence  x,  imaginary,  or  vice  versa. 

But  if  (2)  be  combined  with  the  conjugate  hyperbola, 

a2y2  -  b2x2  =  a2b2, 

•> 

which  is  real, 


b2  -  a2m2 


if  «     .  .  isreal- 

b2  —  a2m2 

Hence   conjugate    diameters   intersect,    one,    the    original 
hyperbola,  the  other,  its  conjugate,  as  aa!  and  W  (Fig.  53) . 

ASYMPTOTES. 

ART.  103.  An  asymptote  of  the  hyperbola  may  be 
defined  as  a  tangent  at  a  point  whose  co-ordinate  are 
infinite,  which,  nevertheless,  intersects  at  least  one  of  the 
co-ordinate  axes  at  a  finite  distance  from  the  origin. 

To  find  the  equation  of  the  asymptotes  then,  it  is  neces- 


142 


Analytical  Geometry. 


sary  to  determine  a  line  that  will  touch  the  hyperbola  at 
infinity  (Fig.  54). 


Fig.  54- 

Let  the  equation  of  a  line  be 

y  =  mx  +  c (i) 

and  the  equation  to  the  hyperbola  be 

b~x2  —  a2y2  =  a2b2 (2) 

Combining  (i)  and  (2), 

b2x2  —  a2m2x2  —  2  a2mcx  —  a2c2  =  a2b2, 
or        x2  (b2  —  a2m2)  —  2  a2mcx  —  (a2c2  +  a2b2)  =  o 
wherein  the  values  of  x  are  the  abscissas  of  the  point  of 
intersection.     By  the  theory  of  equations,  these  values  will 
be  infinite  if  the  coefficient  of 


that  is,  if 
or 


x2=  o, 
b2  —  a2m2  =  o 


m=  ±-  - 
a 


Analytical  Geometry.  143 

For  in  the  typical  quadratic,  ax2  +  bx  +  c  =  o 


_  -  b  +  V  b2  -  4  ac  -  b  -  Vb2  -  4  ac 

^    — — ^— — — _— — —  QJ"  -      • 

2  a  2  a 

In  either  case  if  the  denominator  2  a  =  o  or  a  =  o  the 
values  of  #  will  be  infinite,  having  a  -denominator  o;  but  a 
is  the  coefficient  of  x2;  hence  the  rule. 

/.  if  m  =  ±  —  the  line  y  =  mx  +  c  meets  the  hyperbola 

b2x2  —  a2y2  =  a2^2  at  infinity. 

We  found,  however,  in  Art.  107,  that  the  slope  equation, 
of  the  tangent  to  the  hyperbola  is, 

y  =  mx  ±  Va2m2  —  b2; 


that  is,  in  y  =  mx  +  c,  if  c  =  ±Va2m2  —  b2,  y  =  mx  +  c 
becomes  a  tangent. 

If  m  =  ±  —  ,  however, 
a 

/T27l2 

a2m2  -  b2=  ~  -  b2  =  b2  -  b2  =  o. 
a2 

/.  at  infinity  y  =  mx  +  <:  becomes  a  tangent  if  c  =  o 

and  m  =  ±  —  •        Hence  the  equation  to  an  asymptote  is 
a 

b  b 

y  =  —x    or    y— x. 

a  a 

The  form  of  these  equations  shows  that  the  asymptotes 
pass  through  the  origin. 

ART.  104.  Relation  between  the  equations  of  the  asymp- 
totes and  that  of  the  hyperbola. 

Clearing  the  two  above  equations  of  fractions,  trans- 
posing and  multiplying  together, 

(ay  —  bx)     (ay  +  bx)  =  o, 
or  a2y2  —  b2x2  =  o    or    b2x2  —  a2y2  =  o. 


144  Analytical  Geometry. 

Comparing  this  with  b2x2  —  a2y2  =  a2b2,  it  is  observed  that 
they  are  the  same  except  for  the  constant  term  a2b2,  hence 
given  its  two  asymptotes  it  is  easy  to  write  the  equation  of 
the  hyperbola,  or  vice  versa. 

If  y  =   —  x  and  y  =  —    —  x  are  the    equations  of   the 
a  a 

asymptotes  to  a  hyperbola,  its  equation  may  be  written, 

b2x2  -  a2y2  ±  C  =  o (») 

the  minus  sign  of  C  indicating  the  primary  hyperbola;,  the 
plus  sign,  its  conjugate.  If  in  addition  a  point  is  given 
through  which  the  hyperbola  must  pass,  C  can  be  deter- 
mined. 

For  example :  The  asymptotes  of  a  hyperbola  are  y  =  J  x 
and  y  =  —  J  x.  If  the  hyperbola  passes  through  the 
point  (6,  2Va),  to  find  its  equation.  The  equation  will  be 

(2  y  —  x)  (2  y  +  x)  ±  C  =  o 
or  4  y2  -  x2  ±  C  =  o. 

Substituting; 

4  (2 VI)2-  (6)2±C=o, 

whence  C  =  ±  4,  whence  4  y2  —  x2  ±  4  =  o  are  the  equa- 
tions to  primary  and  conjugate  hyperbola. 

Corollary:  The  same  principle  will  clearly  apply  no  matter 
where  the  origin  is  taken,  since  both  hyperbola  and  asymp- 
totes are  referred  to  the  same  point  as  origin,  and  hence 
the  relation  between  their  equations  remains  the  same. 
For  example,  if  2  y  —  3  x  —  i  =  o  and  ;y  +  2jc  +  3=o, 
are  the  asymptotes  of  a  hyperbola,  its  equation  is, 

(y  +  2X  +  3)   (2  y  -  3  x  -  i)  ±  C  =  o. 

ART.  105.  It  is  often  desirable  to  refer  the  equation  of  a 
hyperbola  to  its  asymptotes  as  axes. 


Analytical  Geometry. 


'45 


By  determining  the  angles  made  by  the  new  axes  (the 
asymptotes)  and  the  old,  and  using  the  transformation 
equations  (J'),  Art.  38,  the  result  is  most  readily 
achieved. 

These  equations  are 


y  =  x'  sin  6  +  /  sin  0)  ) 

X  =   X'  COS   0  +  /  COS  (f>)  > 


0  -  reflex  ZXON  =  Z  -XON, 


.,,. 
MOX     (Fig.  55). 


Fig.   55- 


Since   the   new   axes   are    asymptotes,   their  slopes  are 

•\ —  and  —  —  from  their  equations,  that  is, 
a  a 


tan  0  =  -     -  ; 

a 


tan  <j)  =  — 
a 


146  Analytical  Geometry. 

whence  by  Goniometry, 


b 


COS0=  -=4 


+  ft2  '  Va2  +  ft2  ' 


sin  d>  =  -  .  cos  d>  = 


a 


'a2  +  b2  Va2  +  ft2 

Substituting  these  values  in  (J'), 
ft 


Va2  +  ft- 


(/-*')      ....     (i) 


*  = 


Substituting  (i)  and  (2)  in  the  hyperbola  equation, 
ft2*2  -  a2y2  =  a2b2, 


or  (/  +  ^)2  -  (/  -  y)2  =  a2  +  b2, 

whence  4  ^y  =  a2  -f-  ft2. 

Dropping  accents, 

4*y=a2  +  b2=c2    .     .     .     .     (Aa>  h) 

which  is  the  equation  of  a  hyperbola  referred  to  its  asymp- 
totes. 

It  shows  that  the  co-ordinates  of  a  hyperbola  referred  to 
its  asymptotes  vary  inversely  as  one  another. 

ART.  106.  Equation  of  the  tangent  to  the  hyperbola 
referred  to  its  asymptotes. 

Pursuing  exactly  the  same  method  as-  before,  we  deter- 
mine the  equation  of  a  secant  line  and  revolve  this  line  to  a 
tangent  position. 


Analytical  Geometry.  147 

The  equations  of  any  line  through  (x',  /)  and  (x",  y")  is 


If  the  points  (x',  y')  and  (x",  /')  are  on  the  hyperbola, 
they  must  satisfy  4xy=  c2. 

.-.  4X'y'  =  c2    ......  (i) 

4x"y"=  c2    ......  (2) 

Subtracting  (i)  from  (2)  and  simplifying; 

x"y  -  x'y'  =  o    or    x"f  =  x'y'  ...  (3) 


Subtracting  x"yf  from  both  sides  to  get  the  value  of  2 

x"— 


x"y"  —  x"yf  =  x'y'  —  x"yf. 
Factoring;         x"  (y"  —  y')  =  —  y'  (x"  —  x') 


or  »•  -  ^-  =  —  -^—  . 

x"  -  x'  x" 

Substituting  in  B, 
y  —  y  =  -     2—    (x  —  x')  (4).     [The  equation  of  a  secant.] 

OC 

As  the  points  approach  coincidence  y?  approaches  x' 
and  y  approaches  /,  and  eventually  x"  =  x',  y"  =  yf. 
Substituting  in  (4); 

y  -  yf  =  -  ^  (x  -  x') 

whence  x'y  —  x'y'  =  —  xyf  + 

x'y  +  xyr  —  2  x'y', 

y-,  +  ^=* 
/     v 


148  Analytical  Geometry. 

EXERCISE. 

Tangents  and  Asymptotes. 

Find  the  equation  of  a  tangent  to  the  following  hyper- 
bolas: 

1.  2  x2      •    3  y2  =  12,  at  (12,  2)_._ 

2.  16  y2  -  •    9  #2  =  144,  at  (4  \/3,  6). 

3.  x2  4/=  4  at  (?,  |). 

4.  i6#2  -      9y2=  J44  at   (?,  3). 

5.  25  /  --  16  x2  =  400  at  (3!,  ?). 

6.  36  y2  -  25  x2  =  goo  at  (3^,   ?). 

7.  Find  the  normal  to  each  of  the  above. 

8.  What  points  on  a  hyperbola  have  equal  subtangent 
and  subnormal  ? 

9.  What  are  the  equations  of  the  tangents  to  the  hyper- 
bola 16  x2  —  9  y2  =  144,  parallel  to  the  line  3?— 5^  +  3=0? 

10.  What  are  the  equations  of  the  tangents  to  the  hyper- 
bola x2  —  4  y2  =  4,  perpendicular  to  the  line  y  =  -2^  +  3? 

11.  What  is  the  equation  of  the  normal  to  the  hyperbola 
x2  —  4  y2  =  4,  perpendicular  to  the  line  y  =    -  2  #  +  3  ? 

12.  Find    the    equations  of    the    common    tangents    to 
16  x2  —  25  y2  —  400  and  x2  +  y2  =  9. 

13.  Find  the  slope  equation  of  a  tangent  a2y2  —  b2x2  = 
a2b2. 

14.  Find   the   equations    of   tangents   to   the   hyperbola 
2  x2  —  y2  =  3,  drawn  through  the  point  (3,  5). 

15.  Find  the  equations  of  tangents  drawn  from  (2,  5)  to 
the  hyperbola  16  x2  —  25  y2  =  400. 

16.  Find  the  equations  of  the  tangents  to  the  hyperbola 
1 6  y2  —  9  x2  =  —  144,    which    with    the    tangent    at    the 
vertex  form  an  equilateral  triangle. 

17.  Find  the  angle  between  the  asymptotes  of  the  hyper- 
bola 16  x2  —  25  y2  =  400. 


Analytical  Geometry.  149 

18.  What    is    the    equation    of    the    hyperbola    having 
y  —  2  x  +  i  =  o  and  3  x  +  3  y  —  5  =  o  for  its  asymp- 
totes, if  it  passes  through  (o,  7)? 

19.  Show  that  the  perpendicular  from  the  focus  of  a 
hyperbola  to  its  asymptote  equals  the  semi-conjugate  axis. 

20.  Find  the  equations  of  the  tangents  to  the  hyperbola 
9  y2  —  4  x2  =  56  at  the  points  where  y  —  x  =  o  intersects  it. 

21.  A  tangent  to  the  hyperbola  9  x2  —  25  y2  =  225  has 
the  ^-intercept  =  —  3.     Find  its  equation. 

22.  Two  tangents  are  drawn  to  9  x2  —  4  y2  =  36  from 
(i,  2).     Find  the  equation  of  the  chord  joining  the  points 
of  contact. 

23.  The  product  of  the  distances  from  any  point  on  a 
hyperbola    to   its    asymptotes   is    constant.     What   is    the 
constant  ? 

24.  Show  that  the  sum  of  the  squares  of  the  reciprocals 
of  the  eccentricities  of  conjugate  hyperbolas  equals  unity. 

25.  The    equation    of    a    directrix    of    the    hyperbola 
b2x2  -  a2y2  =  a2b2,    being 

x  =   ^1.      [c  =  Va2  +  b2], 
c 

show  that   the  major  auxiliary  circle  passes  through  the 
points  of  intersection  of  the  directrix  with  the  asymptotes. 

ART.  107.     Supplemental  chords. 

Supplemental  chords  in  the  hyperbola  are  denned  as 
they  were  in  the  circle  and  ellipse,  hence  from  the  relation 
between  ellipse  and  hyperbola  the  relation  between  the 
slopes  of  supplemental  chords  in  the  hyperbola  is, 

b2 
mm'  =  — -      [putting  —  b2  for  b2  in  ellipse  condition]. 

Since  this  is  also  the  relation  between  the  slopes  of  conjugate 


15°  Analytical  Geometry. 

diameters,  it  follows  that  there  is  a  pair  of  diameters  parallel 
to  every  pair  of  supplemental  chords,  which  suggests  an 
easy  method  of  drawing  conjugate  diameters. 

ART.  1 08.     The  eccentric  angle. 

Since  the  ordinates  of  the  hyperbola  do  not  cut  the 
auxiliary  circles,  the  eccentric  angle  of  a  point  is  not  so 


Fig.  56- 

readily  determined  as  in  the  ellipse  and  a  more  arbitrary 
definition  is  necessary.     The  angle  </>  so  determined  that 

x  =  a  sec  <j)  and  y  =  b  tan  (j>, 

is  called  the  eccentric  angle  for  the  point  (x,  y).     These 
values  will  satisfy  the  equation 


for  substituting; 

a2b2  sec2  <    - 


tan 


a>b\ 


Analytical  Geometry.  151 

or  sec2  <fi  —  tan2  ^>  =  i. 

which  is  true  by  goniometry. 

To  construct  this  angle  for  a  given  point,  the  auxiliary 
circles  [with  radii  a  and  b]  are  drawn.  (Fig.  56.) 

Let  P  be  any  point  on  the  hyperbola.  Draw  its  ordinate 
PD  and  from  the  foot  of  PD  draw  a  tangent  to  the  major 
auxiliary  circle  touching  it  at  C,  then  Z.  COD  =  <£  for 
point  P,  (xt  y). 

For,  draw  BE  a  parallel  tangent  to  the  minor  circle,  then 
in  the  right  triangle  OCD, 

cos  COD  -          '-  =  -  [OD  =  abscissa  of  P] 

or  x  =  a  sec  COD (i) 

Again  in  the  right  triangle  OBE 

tan  BOE  =  tan  COD  =  —  (2) 

OB 

The  triangles  COD  and  BOE  are  similar. 

.'.  OB  :  OC  :  :  BE  :  CD, 

whence 


BE  =   OB  x  CD  =   OB  VOD2  -  OC2        

OC  OC  a 

or  BE2  =    ¥-  (x2  -  a2). 

az 

But       f  -    -2   (x2  -  a2)  from    (AA  ).     .-.  BE  =  y. 

Hence  from  (2)  tan  COD  =  y- 

b 

or  y=  b  tan  COD       ...     (3) 

Comparing   (T)   and    (3)   with  the  condition  equations 
for  0,  we  see  that  COD  =  <f>. 

Hence   the   eccentric   angle  is  found   by  drawing  from 


152  Analytical  Geometry. 

the  foot  of  the  ordinate  of  a  point,  a  tangent  to  the  major 
auxiliary  circle.  Then  the  angle  formed  with  the  axis  by 
the  radius  drawn  to  the  point  of  tangency  is  the  eccentric 
angle  for  that  point.  The  eccentric  angle  is  used  to  best 
advantage  in  the  calculus. 

ART.  109.  There  are  two  interesting  geometrical  prop- 
erties of  the  hyperbola  when  referred  to  its  asymptotes. 

(a)  The  product  of  the  intercepts  of  any  tangent  on  the 
asymptotes  is  the  same. 


Fig.  57. 


Let  BPC  (Fig.  57)  be  a  tangent  at  P,  then  its  intercepts 
on  OX  and  OY  (OB  and  OC),  respectively,  will  be  found 
by  setting  successively  y  =  o  and  x  =  o  in  its  equation, 


^    +-=2, 
„,'     '    «S  ' 


whence 
and 


x 


=  OB  =  2  yf  ) 

,       (*',  /  being  point  P), 

y  =  OC  =  2  /  ) 


Analytical  Geometry.  153 

multiplying;     OB  .  OC  =  4  x'  y'  =  a2  +  b2    (a  constant). 
Since  x'y'  is  on  the  hyperbola  4  x'y'  =  a2  +  62. 

(6)  77ze  araz  <?/  //ze  triangle  formed  by  a  tangent  and  the 
asymptotes  is  constant.  The  area  of  the  triangle  BOC 
(Fig.  57),  by  trigonometry,  is 

Area  BOC  =  ^OC  gin     BQC  =  OROC  ^ 

2  2 

[COA  -  BOA  =  0,  Art.    105]  -  OB  .  OC  sin  $  cos  0 
[since  sin  2  0  =  2  sin  <p  cos  0  ]  — 

OB.OC.  ~==.       .    a         =QB.OC 


But  OB  .  OC  =  a2  +  b2. 

.;.  area  BOC  =  (a2  + 

That  is,  the  area  of  this  triangle  always  equals  the  product 
of  the  semi-axes. 


EXERCISE. 
General  Examples. 

1.  If     ^=3^  +  15     is     a     chord     of     the    hyperbola 
36  x2  —  16  y2  =  576,  what  is  the  equation  of  the  supple- 
mentary chord  ? 

2.  The  point  (5,  f )  lies  on  the  hyperbola  4  #2  —  9  ^=36. 
Find  the  equations  of  the  diameter  through  this  point  and 
of  its  conjugate. 

3.  Find  the  equation  of  the  line  passing  through  a  focus 
of  a  hyperbola  and  a  focus  of  its  conjugate  hyperbola. 

4.  Find  the  angle  between  a  pair  of  conjugate  diameters 
of  the  hyperbola,  b2x2  —  a2y2  —  a2b2. 


154  Analytical  Geometry. 

5.  Find   the   equation   of   the   chord   of   the   hyperbola 
9  x2  —  1 6  y2  =  144,  which  is  bisected  by  the  point  (2,  3). 

6.  Show  that  the  locus  of  the  vertex  of  a  triangle,  whose 
base  is  constant,  and  the  product  of  the  tangents  of  its  base 
angels  is  a  negative  constant,  is  a  hyperbola. 

7.  Show  that  the  eccentric  angles  of  the  extremities  of 
a  pair  of  conjugate  diameters  are  complementary. 

8.  What  is   the   equation   of  the  focal  chord  which  is 
bisected  by  the  line  y  =  6  x  ? 

9.  In   the   hyperbola   9  x2  —  16  y2  =  144,   what   is   the 
equation  of  the  diameter  conjugate  to  y  —  3^=0? 

10.  Show  that  tangents  at  the  ends  of  conjugate  diam- 
eters intersect  on  the  asymptotes. 

11.  The  base  of  a  triangle  is  2  b  and  the  difference  of 
the  other  sides  is  2  a.     Show  that  the  locus  of  the  vertex  is 
a  hyperbola.     [Take  the  middle  of  the  base  as  origin.] 

12.  For  what  point  of  the  hyperbola  xy  =  12  is  the  sub- 
tangent  =  4  ? 

13.  Show  that  an  ellipse  and  hyperbola  which  have  the 
same  foci  intersect  at  right  angles. 

14.  What  are  the  equations  of  the  tangents  to  the  hyper- 
bola x2  —  4  y2  =  4,  which  are  perpendicular  to  the  asymp- 
totes ? 

15.  In    the    hyperbola    25  x2  —  16  y2  =  400,    find    the 
equations   of   conjugate  diameters  that    cut   at   an    angle 
of  45°- 

1 6.  In    the    hyperbola    16  x2  —  25  y2  =  400,    what    are 
the  co-ordinates  of  the  extremity  of  the  diameter  conjugate 
to  25  y  -f  16  x  =  o? 

17.  In  the  hyperbola  4  x2  —  9  y2  =  36,  the  equation  of  a 
diameter  is  3  y  —  2  x  =  o.     What  is  the  equation  of  any 
one  of  its  system  of  chords  ? 


CHAPTER   VIII. 
HIGHER   PLANE    CURVES. 

ART.  101.  There  are  several  other  curves  known  as 
Higher  Plane  Curves  because  their  equations  are  more 
complex,  that  are  used  extensively  in  engineering.  These 
we  will  consider  briefly. 

THE    CYCLOID. 

The  cycloid,  much  used  in  gear  teeth,  is  the  curve  gener- 
ated by  a  point  on  the  circumference  of  a  circle  of  given 
radius,  as  the  circle  rolls  along  a  straight  line.  The  circle 
may  be  called  the  generator  circle,  and  the  straight  line  the 
directrix. 


Fig.   58. 


To  f,nd  its  equation.  Let  P(  Fig.  58)  be  the  generating 
point,  r  the  radius  CP,  OE  =  x  and  PE  =  y  for  P,  and 
call  Z  PCB,  6. 

Then  PE  =  CD  -  CB  =  r  -  r  cos  6. 


156  Analytical  Geometry. 

That  is,  y  =  r  —  r  cos  0  ..........     (i) 

Also       x  =  OE  =  OD    -  -   ED  =  OD  -  PB   =  rO  - 
r  sin  6     ...............     (2) 

Since  6  is  an  extra  variable,  its  elimination  is  necessary. 


From  (i)        cos  6  =  r- 


-^-  =  i  -  y-  , 


whence 

i  —  cos  6  =  vers  0  =  —    or     6  —  vers"1  *-  . 
r  r 

Substituting  this  value  of  6  in  (2), 

x  =  r  vers"1  ^-  —  r  sin  (  vers"1  £  ] 
r  \  r) 

or  x  —  r  vers"1  —  —  \72  ?y  —  y2. 

For  vers"1^-=  6, 

r 


=  vers  0=i  —  cos  6, 


=  cos*  0. 


_  /JlrJ!  J_ 


Whence  an  fl  - 


and  r  sin  ^  =  r  sin  j  vers"1  —  ) 


V  2  r    —    2. 


Analytical  Geometry.  157 

CONSTRUCTION    OF    THE    CYCLOID. 

ART.  in.  From  the  nature  of  the  development  of  the 
cycloid,  it  is  readily  constructed  by  points.  The  first 
method  to  be  shown  produces  an  accurate  cycloid  if  suffi- 
cient points  be  taken. 

The  second  method,  which  is  employed  in  mechanical 
drawing,  gives  a  cycloid  of  sufficient  approximation. 

First  Method :  Let  M  be  the  generator  circle  in  its 
middle  position,  and  XX'  the  directrix.  Make  OV  equal 
i  the  circumference  of  M.  Divide  the  semi-circumference 
OCN  into  6  equal  parts,  also  OV  into  6  equal  parts.  Then 


V^^/~K   ^\""\ 

«/-           h                  \           X 

B/ 

IB  M    y        \ 

V 

\A              y7                               \ 

X'.       'ill 

1      r^^~—  i  —  ""^                                    "^r 

Fig.  59- 

clearly  the  6  points  on  OCN  would  exactly  coincide  with 
the  6  points  on  OV  if  the  circle  were  rolled  back  toward  V. 
Through  the  division  points  on  OCN:  A,  B,  C,  D,  E, 
draw  lines  parallel  to  the  directrix.  Now  if  the  circle  were 
revolved  toward  V  until  A  and  P  coincided,  then  N  would 
be  on  the  level  now  occupied  by  E,  that  is,  it  would  be 
somewhere  on  the  parallel  through  E;  N  would  still  be  the 
same  distance  from  A  that  it  now  is;  hence  if  we  take  a 
radius  AN,  with  P  as  a  centre,  we  will  cut  the  parallel 
through  E  in  the  place  where  N  was  when  A  was  at  P. 
Likewise  with  Q  as  a  centre  and  radius  BN,  cut  the  parallel 
through  D,  and  we  have  the  position  of  N  when  B  was  at 
Q.  The  same  process  continued  will  give  all  the  succes- 


Analytical  Geometry. 


sive  positions  of  N,  and  if  these  be  joined  by  a  smooth 
curve,  we  have  the  cycloid  described  by  N. 

ART.  112.  Second  Method  :  This  approximate  construc- 
tion used  in  mechanical  drawing  is  based  on  the  fact  that 
for  very  small  arcs  the  arc  does  not  sensibly  differ  from  its 
chord,  so  the  divisions  are  "  stepped  off  "  with  the  com- 
passes, thus  really  getting  chords  not  arcs,  but  by  taking 
the  distances  small  enough,  any  degree  of  approximation 
may  be  attained. 

Draftsmen  use  this  slightly  modified  method,  which 
gives  a  sufficient  approximation,  as  follows: 


E' 


ABODE 


Fig.  60. 


Fig.  60.  Let  MN  be  the  directrix  and  C  the  generator 
circle.  Lay  off  any  small  distance  on  MN  a  sufficient 
number  of  times  choosing  the  distance  small  enough  so  that 
as  a  chord  it  would  not  sensibly  differ  from  its  arc,  as  AB. 
Then  AB,  BC,  CD,  etc.,  will  practically  equal  corresponding 
arcs  on  C.  Draw  a  series  of  circles  (or  parts  of  them) 
having  the  radius  of  C.  These  represent  the  generator 
circle  in  its  successive  positions. 

From  B,  C,  D,  etc.,  successively  "  step  off  "  with  com- 
passes on  the  arc  passing  through  them,  i,  2,  3,  etc.,  units 
(as  AB).  These  will  give  points  on  the  cycloid  as  A',  B', 
C',  D',  etc.  The  curve  drawn  through  these  points  will  be 
a  very  good  approximation. 


Analytical  Geometry. 


ROULETTES. 


'59 


The  hypocydoid  is  described  by  a  point  on  the  circum- 
ference of  a  circle,  which  rolls  on  the  inner  side  of  the 
circumference  of  a  second  circle. 

If  the  generator  circle  rolls  on  the  outside  of  the  circum- 
ference of  the  directrix,  the  resulting  curve  is  called  an 
epicycloid. 

The  two  circles  may  have  any  relative  radii,  and  if  the 
ratio  between  them  is  commensurable,  the  cycloids  will  be 
closed  curves,  consisting  of  as  many  arches  as  the  ratio  con- 
tains units.  The  common  ratio  is  4.  If  the  ratio  is  i,  the 
epicycloid  resulting  is  called  a  cardioid  (see  Art.  16). 

Curves  described  by  rolling  one  figure  upon  another  are 
known  collectively  as  roulettes. 

ART.  114.     To  find  the  equation  of  the  hypocydoid. 

Let  circle  C  be  the  directrix  and  circle  C'  the  generator 
circle  (Fig.  61).  Let  P  be  the  generating  point,  starting 


Fig.  61. 

from  coincidence  with  D.  Draw  the  co-ordinates  of  P, 
CF  and  PF  (x,  y)\  C'  E  perpendicular  to  CD  and  PA  ||  to 
CD,  and  let  CD  and  CY  (_[_  to  CD  through  C)  be  the 
axes.  Let  Z  BCD  =  </>,  Z  BC'P  =  «;  Z  C'PA  -  0; 
CB  =  r  and  C'B  -  /. 


160  Analytical  Geometry. 

Then   CF  =  CE  -  FE  =  CE  -  PA  =  CC'  cos   <f>  - 
C'P  cos  6  or  x  =  (r  -  r')  cos  ^  -  /cos  d.    .     .     .     (i) 
Extend   C'P   to    meet    CD    at   G;    Z  C'GD  =   6,   and 
a  =  <j>  +  C'GC  =  (/>  +  (180  -  0) 

[a  is  exterior  angle  of  triangle  C'GC]. 
Hence  a  —  (f>  =  180  —  6. 

cos  (a  —  (f>)  =  cos  (180  —  0)'=  —  cos  0     [Goniometry]. 
Substituting  in  (i); 

x  =  (r  —  /)  cos  <j)  -f  r'  cos  (a  —  </>)    .     .     (2) 

Likewise,     y  =  (r  —  r')  sin  (j>  —  r'  sin  (a  —  0)    .     .     .(3) 
But  since  arc  BD  =  arc  BP  by  method  of  descrip- 
tion of  the  hypocycloid  rcj)  =  r'  a,  or  a  =    ,    • 

Substituting  in  (2 )  and  (3 ) ; 

x  =  (r  -  /)  cos  $  +  /  cos   (r  .     .     (a) 

y  =  (r  -  /)  sin  0  -  /  sin    (r  ~  ^  ^     .     .     (b) 

If  <^>  be  eliminated  between  (a)  and  (ft)  the  rectangular 
equation  for  the  hypocycloid  results,  but  in  this  general 
form  the  equation  would  be  exceedingly  complicated. 

But  if  r  =  4  r',  as  is  customary,  the  result  is  compara- 
tifely  simple,  thus: 

(a)  becomes;  x  =  }  r  cos  </>  +  |  r  cos  3  <^>. 

(b)  becomes;  y  =  f  r  sin  <yS  —  J  r  sin  3  c/>, 

or  #  =  -  (3  cos  (/>+  cos  3  <£)     .     .     (a') 

4 

and  y  =  -  (3  sin  <£  —  sin  3  0)      .     .     (6') 

4 

T,     ^  .  ?  ?  cos  (/>  +  cos  3  (£  =  4  cos3  </> 

By  Tngonometry  >  «*-  f  • 

)  3  sin  9  —  sin  3  0  =  4  smd  (p 


Analytical  Geometry. 


161 


Hence  (a'}  becomes  x  =  r  cos3  <£    .     (a") 

and  (bf)  becomes  y  =  r  sin3  <j>     .     (b") 

Combining  (a"}  and  (&");  #*  =  r*  cos2  <£, 


Add;  #3  +  y$  =  r3  [since  cos2<£  +  sin2<£  =  i]. 

ART.  115.     To  construct  the  hypocydoid. 

Let  C  be  the  directrix;  (Fig.  62)  C'  the  generator  circle; 
P  the  generating  point.  Divide  the  quadrant  P'K  into  8 
equal  parts  and  the  semicircle  PE'  into  4  equal  parts.  Let 
P  start  at  P',  then  when  A'  and  A  coincide  as  the  circle  C' 


Fig.  62. 

rolls,  P  will  be  at  the  distance  DD'  from  P'  and  at  the  dis- 
tance AT  from  A.  Hence  with  Px  as  a  centre  and  DD' 
as  radius  describe  an  arc  intersecting  another  described 
with  A  as  centre  and  A'P  as  radius.  This  intersection 
point  will  be  a  point  on  the  hypocycloid. 

When  B'  is  at  B,  P  will  be  at  the  distance  BB'  from  P' 
and  at  the  distance  B'P  from  B.  The  intersection  of  arcs 
described  with  centres  Pr  and  B  and  radii  BB'  and  B7P, 
respectively,  will  be  a  second  point  on  the  hypocycloid, 
and  so  on. 


1 62  Analytical  Geometry. 

Evidently  the  greater  the  number  of  equal  parts  into 
which  the  quadrant  and  the  generator  circle  are  divided 
the  more  accurate  will  be  the  hypocycloid. 

If  the  ratio  of  the  radii  of  the  two  circles  is  3,  the  entire 
directrix  will  be  divided  into  3  times  as  many  parts  as  the 
circumference  of  the  generator  circle  and  similarly  for  any 
ratio.  In  the  figure  62  the  ratio  is  4. 

ART.  1 1 6.  Draftsman's  method  of  constructing  the  hypo- 
cycloid. 

This  method  is  -almost  exactly  similar  to  that  described 
for  the  cycloid,  using,  however,  angular  division  of  the 
directrix,  which  is  now  a  circumference. 


Fig.  63. 

Fig.  63.  Let  C  be  the  centre  of  the  directrix  and  C'  the 
generator  circle.  "  Step  off  "  on  the  circumference  of  C 
any  small  equal  arcs  as  AB,  BD,  DE,  etc.;  at  A,  B,  D,  etc., 
draw  tangent  circles  equal  to  C7.  From  A,  B,  C,  D,  E,  etc., 


Analytical  Geometry.  163 

successively  "  step  off  "  i,  2,  3,  4,  etc.,  times  the  distance 
AB,  the  resulting  points  will  determine  the  hypocycloid. 
An  exactly  similar  process  will  produce  the  epicycloid,  if 
the  generator  circle  be  rolled  on  the  outside. 

ART.  117.  Another  form  of  roulette  is  the  involute, 
which  is  described  by  a  fixed  point  on  a  straight  line,  that 
rolls  as  a  tangent  on  a  fixed  circle.  Let  C  (Fig.  64)  be  the 
directrix  circle  and  MN  the  initial  position  of  the  line. 


Fig.  64. 

"  Step  off  "  any  small  equal  arcs  on  the  circumference  of 
C  as  AB,  CD,  DE,  etc.  Draw  tangents  at  the  points  of 
division  and  beginning  with  A  stepoff,  successively  i,  2, 
3,  4,  etc.,  times  the  distance  AB  on  the  tangent  lines.  The 
resulting  points  will  determine  an  involute.  Any  curve 
whatever  will  produce  an  involute  in  this  way,  but  the 
circle  is  most  commonly  used.  A  gear  tooth  is  made  up 
of  cycloid,  evolute,  and  circular  arc  in  varying  proportions. 

SPIRALS. 

ART.  118.  A  spiral  is  described  by  a  point  receding, 
according  to  some  fixed  law,  along  a  straight  line  that 
revolves  about  one  of  its  points.  There  are  a  number  of 


164 


Analytical  Geometry. 


spirals,  one  of  which  will  illustrate  this  type  of  curve.  The 
revolving  line  is  called  the  radius  vector  and  the  angle  it 
makes,  in  any  position,  with  the  initial  line,  is  called  the 
rectorial  angle. 

The  hyperbolic  spiral  is  the  curve  generated  by  a  point, 
which  moves  so  that  the  product  of  radius  vector  and 
vectorial  angle  is  constant. 


Fig.  65. 

Calling  the  radius  vector,  r ;  the  vectorial  angle  6  and  the 
constant  C,  we  have  by  definition, 


r6=  C. 


ii 


To  construct  it  when  C  =  n,   then  r  =  — 

a 


Analytical  Geometry.  165 


Make  a  table  of  values  for  r,  as  follows; 
When  6  =  o,  r  =  oo  ,    TT  =  3  \. 

0=    ->     (45°),  'H    I4- 

4 

6  =    -,    (60°),  r  =  10.5. 
0  -  "S-5  ,  (75°),  r  -  8.4. 


=  7~,    .(90°),  r-7- 


r=45,etc. 


One  complete  revolution  of  the  radius  vector  from  o°  to 
360°  describes  a  spire,  as  from  GO  to  B  [Fig.  65],  and  the 
circle  described  with  the  final  radius  vector  of  the  first 
spire,  as  radius,  is  called  the  measuring  circle, 


ELEMENTARY    CALCULUS. 


ELEMENTARY   CALCULUS. 


CHAPTER  I. 
FUNDAMENTAL   PRINCIPLES. 

ART.  i.  Variables  and  constants.  Suppose  we  wish  to 
plot  a  curve,  corresponding  to  the  relation  y  =  x3  +  2  x2 
-  $  x  —  6;  and  for  this  purpose  assign  to  x  certain  arbi- 
trary values,  calculating  from  these  the  corresponding  and 
dependent  values  of  y.  Now  in  such  a  case  both  x  and  y 
are  variable  quantities,  x  being  called  an  independent, 
and  y  a  dependent  variable. 

In  general:  A  Variable  is  a  quantity  which  is  subject  to 
continual  change  of  value,  while  an  Independent  Variable 
is  supposed  to  assume  any  arbitrary  value,  and  a  Depen- 
dent Variable,  is  determined  when  the  value  of  the  Inde- 
pendent Variable  is  known. 

Examples  :    y  =  x*,     y=  tan  x,     y  =  log  x. 
In  the  above  examples  x  is  the  independent,  and  y  the 
dependent  variable. 

When  a  quantity  does  not  change  or  alter  its  value  such 
as  TT  =  3.14159  .  .  .  ,  it  is  called  a  Constant  Quantity,  or 
simply  a  Constant. 

ART.  2.  Functions.  Let  us  again  take  the  equation 
y=x3  —  4xz+x  +  6')  we  know  that  for  every  value 
of  x  there  is  a  corresponding  value  of  y;  not  necessarily 
different,  for  if  x  =  3,  y  =  o,  and  if  x  =  2,  y  =  o,  but 

169 


170  Elementary  Calculus. 

nevertheless  to  each  value  of  x  there  corresponds  a  certain 
definite  value  of  y.  When  two  quantities,  x  and  y,  are 
related  in  this  manner  we  say  that  y  is  a  /unction  of  x. 

In  the  examples  given  above,  namely,  y  =  tan  x,  y  =  x4y 
y  =  log  x,  we  see  that  in  each  case  if  we  assign  a  value 
to  x  there  corresponds  a  definite  y  value;  we  therefore  call 
y  a  function  of  x. 

Again,  if  we  note  the  barometer  readings  corresponding 
to  each  hour  of  the  day,  we  can  involve  the  observations 
in  a  curve,  and  we  say  that  the  height  of  the  barometer  is  a 
function  of  the  time,  because  to  each  change  in  the  time 
there  corresponds  a  certain  definite  barometric  height. 
It  is  equally  true  that  the  barometer  readings  are  a  func- 
tion of  the  time. 

In  general,  A  quantity  P  is  a  junction  of  a  quantity  Q, 
when  to  every  value  which  Q  can  assume  there  corresponds 
a  certain  definite  value  of  P. 

It  is  customary  to  express  the  term  "  function  of  "  by 
the  symbols  F,  /,  <j>  (Phi);  thus  we  write  sin  x  =  F  (x), 
sin  x  =  I  (x}  or,  sin  x  =  (f>  (x),  meaning  that  the  sine  of 
an  angle  is  a  quantity  which  assumes  certain  definite  values 
dependent  upon  the  size  of  the  angle  x.  Again,  if  y  =  cos  x, 
then  y  —  f  (x)  or  in  the  case  of  an  equation  such  as 
y  =  x3  +  2  x*  —  5  x  —  6  we  may  also  write  y=f  (x). 

This  latter  mode  of  expressing  an  equation  briefly  by 
the  symbol  y  =  F  (x)  or  y  =  f  (x}  is  in  very  general  use. 

From  the  definition  of  a  function,  given  above,  we  see 
that  if  an  expression  involves  any  quantity,  it  is  itself  a 

function  of  that  quantity;  for  example,    -^  -  is  a  function  of 

D 

x,  since  this  fraction  has  a  definite  value  corresponding  to 
each  change  in  the  value  of  x,  likewise  3  cos  a  +  5  tan  a 
is  a  function  of  a. 


Elementary  Calculus.  171 

Further,  the  area  of  a  triangle  is  a  function  of  its  base 
and  also  of  its  altitude.  Such  a  double  relation  is  indicated 
thus:  area  A  =  /  (b,  h),  while  the  area  of  a  square  is  a 
function  of  its  side.  If  x  is  a  side  and  y  the  area,  then 
y  =  x2\  we  may  write  this  equation  in  the  general  form 
y  =  f  (x).  Again,  the  volume  of  a  sphere  is  a  function  of 
its  radius,  or  V  =  <f>  (r). 

ART.  3.  Object  of  the  Differential  Calculus.  In  algebra, 
geometry,  and  trigonometry,  the  quantities  which  enter 
into  the  calculations  are  fixed;  they  have  absolute  unchang- 
ing values. 

Now,  suppose  we  wish  to  find  the  greatest  value  that  y 
can  assume,  between  x  =  3  and  x  =  2  when  y  =  x?  —  4  y? 
+  x  4-  6.  Here  we  have  two  variables,  x  and  yt  entering 
into  the  calculation,  each  of  which  may  have  an  infinite 
number  of  values  and  from  which  one  special  value  of  x 
is  sought,  which  is  defined  by  the  condition  imposed. 

A  problem,  such  as  the  above,  involving  the  relation  of 
two  or  more  variable  quantities,  comes  within  the  province 
of  the  differential  calculus.  In  general  the  differential 
calculus  supplies  us  with  a  means  of  obtaining  informa- 
tion regarding  the  properties  of  quantities,  the  number  of 
whose  values  are  infinite,  and  which  vary  according  to 
some  known  law. 

One  of  the  chief  advantages  of  the  calculus  lies  in  the 
comparative  simplicity  with  which  complex  problems 
involving  variable  quantities  are  solved,  problems,  which 
if  attacked  by  other  methods,  would  require  long  and 
tedious  operations  and  sometimes  be  impossible  of  solu- 
tion. 

ART.  4.  The  Differential  Coefficient.  Suppose  an  ob- 
server to  take  notice  of  a  passing  bicyclist,  and  to  estimate 
his  speed  at  10  miles  an  hour;  now,  a  statement  to  this 


172  Elementary  Calculus. 

effect  would  imply  that  the  bicycle  at  the  moment  of  obser- 
vation was  travelling  with  a  velocity,  which  if  maintained 
for  the  next  hour,  would  cause  the  rider  to  cover  10  miles. 
It  does  not  follow,  however,  that  this  will  be  the  case,  for 
5  seconds  later  the  speed  of  the  bicyclist  might  be  either 
reduced  or  accelerated;  further,  the  above  statement  in 
no  way  refers  to  the  velocity  of  the  bicycle  prior  to  the 
time  of  observation,  having  reference  to  the  speed  only,  at 
the  exact  moment  when  the  bicyclist  passed  the  observer. 

Should  it  be  desired  to  make  an  accurate  determination 
of  the  speed  of  the  machine,  we  might  place  two  electrical 
contacts  in  its  path,  which  on  closing  would  cause  the 
time  taken  in  traversing  the  space  between  them  to  be 
automatically  registered.  Then  if  v  =  velocity,  s  =  space, 

/  =   time,  we  have  v  =  —   as  a  measure  of  the  velocity. 

In  choosing  a  position  for  the  second  contact,  we  would 
undoubtedly  select  a  point  near  to  the  first;  because  the 
speed  of  the  machine  at  the  moment  of  passing  the  first 
contact  would  be  unlikely  to  remain  constant  for  a  space 
say  of  100  yards,  but  would  be  less  liable  to  change  in  10 
yards,  less  in  i  yard,  still  less  in  i  foot,  and  so  on. 

Hence  it  is,  that  if  we  wish  to  obtain  an  accurate  result, 
giving  the  velocity  of  a  body  at  the  moment  of  passing  a 
certain  point,  we  measure  as  short  a  portion  of  its  path  as 
is  practicable,  and  divide  by  the  correspondingly  small 
time  interval. 

Let  us  now  examine  a  case  of  uniform  motion;  suppose  a 
point  to  travel  a  distance  of  30  miles  in  6  hours  with  uni- 
form velocity.  Now,  uniform  velocity  implies  that  equal 
lengths  of  path  are  traversed  in  equal  times,  no  matter  how 
small  are  the  time  intervals  considered.  Hence  a  point 
travelling  30  miles  in  6  hours,  at  uniform  speed,  travels 


Elementary  Calculus.  173 

5  miles  in  i  hour,  i  mile  in  one-fifth  of  an  hour,  and  so  on, 
as  indicated  in  the  following  table : 


Space  described 
(in  miles)  . 

Time 
(in  hours). 

Velocity 
(in  miles  per  hour)  . 

30 

6 

~6~ 

=  5 

5 

i 

i 

i 

—     ^ 

i 

i 

5 

i 

.2 

=  5 

i 

10 

i 

So 

V  =                    — 
.02 

==  5 

I 

i 

.OI 

100 

500 

.002 

—  5 

I 

i 

.OOOOO  I 

IOOOOOO 

5000000 

.OOOOOO2 

I 

i 

.OOOOOOOOOOO  I 

1000000000000    5000000000000 


.0000000000002 


Now  it  is  most  important  to  note,  that  no  matter  how 
small  the  space  traversed  may  be,  even  if  beyond  all  possi- 
bility of  measurement  and  conception,  the  ratio  of  any 
such  exceedingly  small  space  to  the  minute  time  interval 
taken  in  traversing  it,  invariably  gives  as  a  quotient  5, 
in  the  example  cited.  The  last  space  taken,  which  is 
.00000000000 1  miles  is  equivalent  to  about  one-six  hundred 
millionth  of  an  inch,  while  the  corresponding  time  interval 
is  .0000000000002  hours,  which  is  approximately  three 

billionths  of  a  second;  the  ratio  -  is   nevertheless  equal  to 
5,  giving  a  velocity  of  five  miles  an  hour. 


174  Elementary  Calculus. 

In  general  we  may  state  that  the  ratio  of  two  quantities, 
each  of  which  is  so  small  as  to  be  entirely  beyond  our  com- 
prehension, may,  nevertheless,  result  in  an  appreciable  and 
practically  useful  quotient,  a  fact  which  should  be  most 
carefully  noted. 

When  we  wish  in  general  to  indicate  that  we  are  consid- 
ering a  small  finite  space,  we  employ  the  symbol  As,  while 

A/  is  used   to   express  a   short   time   interval.     Thus    —^ 

means  that  we  are  comparing  a  small  space  with  a  corre- 
spondingly small  time  interval. 
In  the  example  above,  we  have: 

|£  =5          or          A,  =  5.  A/. 

Carrying  this  conception  still  further  we  may  consider 
As  to  become  smaller  than  any  imaginable  quantity;  in 
other  words,  that  the  space  taken  is  infinitely  small  This 
we  indicate  by  ds,  and  call  ds  a  differential  of  space. 

The  same  process  of  reasoning  applied  to  A 2  gives  dt 
as  representing  an  infinitely  small  time  interval  or  a  differ- 
ential of  time.  We  often  refer  to  ds  and  dt  simply  as  differ- 
entials. The  infinite  reduction  of  the  space  and  time  will 
not  affect  the  value  of  their  ratio.  We  will  still  have 

—  =  5     and  ds  =  5  .  dt. 
dt 

The  value  of  the  ratio  of  two  differentials  such  as  ds 
and  dt,  is  referred  to  by  German  mathematicians  as  a 
differential  quotient;  hence  5,  in  our  case,  is  called  a 
differential  quotient. 

Again,  if  we  write  the  expression,  -^  =  5  in  the  form 
ds  =  5  .  dt,  then  5  becomes  a  coefficient,  for  it  multiplies 


Elementary  Calculus. 


the  differential  of  the  dependent  variable  dt  and  is  there- 
fore called  a  differential  coefficient. 

For  the  present  the  student  might  consider  a  differen- 
tial quotient,  in  general,  as  the  value  of  the  ratio  of  two  differ- 
entials; while  the  term  differential  coefficient  implies  the 
same  quantity  regarded  as  that  factor  oj  the  differential  of 
the  independent  variable  which  makes  it  equal  to  the  differ- 
ential oj  the  dependent  variable. 

It  will  be  found  later  that  these  conceptions  are  suscep- 
tible of  a  deeper  meaning  and  lead  to  results  of  great  prac- 
tical value. 

Progress  in  the  study  of  the  calculus,  primarily  depends 
upon  the  thorough  understanding  of  the  meaning  of  the 
differential  quotient  or  coefficient.  Much  misunderstand- 
ing has  arisen  from  the  fact,  that  when  we  have  such 

expressions  as  above,  viz.    —=5  and  also  ds  =  5.  dt,  it  is 

dt 

customary  to  speak  of  the  5  in  either  case  as  a  differential 
coefficient;  in  the  former  case  it  is  strictly  a  quotient,  which 
quotient  becomes  a  coefficient  when  we  write  ds  =  5.  dt. 

ART.  5.     Rates  of  Increase. 

Suppose  we  have  a  square  Al  (see  Fig.  i),  a  side  of  which 


n 


is  of  unit  length;  further  imagine  that  while  the  left  lower 
corner  remains  fixed,  the  sides  are  capable  of  continuous 


i76 


Elementary  Calculus. 


uniform  extension,  so  that  the  square  Al  assumes  larger 
and  larger  proportions,  thus  passing,  during  this  continuous 
expansion,  through  the  dimensions  shown  by  A2,  A3,  A4, 
in  which  the  side  of  each  new  square  is  one  unit  greater 
than  that  of  the  preceding.  Now  by  an  inspection  of 
AJ,  A2,  A3,  A4,  we  see  that 


Square. 

Side  in  Linear 

Units. 

Area  in  Square 
Units. 

Area  Increase  in 
Square  Units. 

AX 

I 

I 

A 

2 

4 

3 

A3 

3 

9 

5 

A4 

4 

16 

7 

Note  that  if  the  side  of  each  square  is  increased  by  addi- 
tions of  one  linear  unit,  the  area  increases  by  3,  5,  and  7 
square  units,  and  as  the  side  lengthens,  the  greater  is  the 
proportionate  increase  of  area,  in  fact  the  square  might  be 
considered  as  growing  with  an  accellerated  increase  of  area. 
As  before  said  we  are  considering  that  the  square  continu- 
ously expands;  now  in  order  to  compare  the  increase  in 
area  with  the  increasing  length  of  the  side,  we  find  it  con- 
venient to  assume  an  arbitrary  unit  of  time.  Hence  we 
say  the  rate  of  increase  of  the  square  is  greater  than  the  rate 
of  increase  of  its  side. 

This  assumption,  which  is  very  general,  enables  us  to 
compare  the  relative  rate  of  increase  or  decrease  of  any 
two  mutually  dependent  quantities.  Thus  we  say  the  rate 
of  increase  of  the  volume  of  a  sphere,  in  units  of  volume,  is 
greater  than  the  rate  of  increase  of  its  diameter,  in  linear 
units,  and  so  on. 

Let  us  return  to  the  case  of  the  bicycle  and  the  observer 
(Art.  4);  we  found,  that  if  we  wished  to  calculate  the  actual 


Elementary  Calculus.  177 

speed  of  the  bicycle  at  the  moment  of  passing  the  point  fcf 
observation,  then  the  smaller  the  space  measured,  the  more 
accurate  would  be  our  results;  this  would  clearly  hold  if 
the  bicyclist  passed  the  observer  with  an  accelerated  velocity. 

Now  this  case  is  similar  to  that  of  the  square  above  men- 
tioned, for  suppose  the  side  of  the  square,  which  is  con- 
tinuously lengthening,  pass  through  the  point  at  which 
x  =  3  linear  units,  we  might  ask  ourselves,  what  is  the 
relation  of  the  rate  of  increase  of  area  of  the  square,  at  the 
moment  when  x  =  3  to  the  rate  of  linear  increase  of  its 
side. 

Let  the  side  x  =  3  centimetres,  and  let  y  be  the  area  of 
the  square  on  x',  we  thus  get 
y  =  x2  =  9.  Now  let  the  side  x 
receive  a  small  increase,  called 
an  increment,  which  we  will 
represent  by  Ax  (read,  delta  x), 
let  A  x  =0.1  centimeters;  thus  x 
becomes  x+  Ax  =  3  +  0.1  =  3.1. 
Upon  the  increased  side  describe 
a  second  square;  we  now  have 
two  squares  (see  Fig.  2),  and  Fig  2 

the   increase  in    area  of  y,  due 

to  the  increment  A#,  is  represented  by  the  shaded  strip; 
this  increment,  which  we  will  call  Ay,  is  obviously  an 
increment  of  area.  We  thus  have: 


y-x'< 


Area  of  square  on   (x  +  A#)  =  (^  +  A#)2=  (3.i)2=9.6i. 
Area  of  square  on    x  =   x2  =  (3)2    =9- 

Difference  (x  +  kx)2-x2  =  Ay    =0.61. 

Now  the  difference  Ay  =  0.6 1,  is  the  increase  in  area  of 
the  square  y,  in  square  centimetres,  during  the  time 
that  x  increased  from  x=  3  to  x  =  3.1  centimetres;  intro 


178  Elementary  Calculus. 

ducing  the  arbitrary  unit  of  time  before  alluded  to,  we 

say: 

Rate  of  increase  of  square  y  _    0.61  _  Ay  _  , 
Rate  of  increase  of  side  x  .1          Ax 

We  will  now  tabulate  a  number  of  values,  calculated 

exactly  as  above,  for  —  -2  ,  f  or  x  =  3  centimetres: 

Ax 

If  A*  =  o.i  then  4^  =  —  =  6.1 
ax      .1 

A*  =.01         42  =  ^2621=6.01. 

Ax          .01 


A*  =.ooi         42  =  :22622i=  6.001. 

Ax         .001 

A  Ay       .00000060000001       , 

Ax  =  .0000001  —  z  =  -  —=  6.0000001. 

Ax  .000000  1 

We  thus  see  that  —  approaches  the  value  6  more  and 
A* 

more  nearly,  the  less  the  increment  Ax. 

If  Ax  is  infinitely  small,  in  other  words  becomes  the 
differential  dx,  then  the  number  of  zeroes  to  the  right  hand 
of  the  decimal  point  before  the  one  would  be  infinite,  and 
the  value  of  the  quotient  would  be  truly  6.  If  Ax  becomes 
a  differential  of  length,  dx,  then  Ay,  becomes  a  differential 
of  area,  dy,  and  as  the  quotient  6  is  the  result  of  the  com- 
parison of  these  two  differentials,  it  is,  therefore,  a  differen- 
tial quotient;  thus  we  write: 
d 


TT  the  rate  of  increase  of  the  square 

Hence  we  say  -  -  —  —          ~~n  -  ri  -  =  6  at 
the  rate  of  increase  ot  the  side 

moment  when  the   side  is   3   units  in  length.    As  before 


Elementary  Calculus.  179 

mentioned  we  sometimes  write  dy  =  6  dx;  here,  six  figures 
as  the  coefficient  of  the  differential  dx  of  the  independent 
variable,  and  is  therefore  called  a  differential  coefficient.  We 
might  calculate  this  differential  quotient  in  another  man- 
ner, which  would  lead  us  to  a  more  general  result;  thus, 
taking  x  =  3,  and  .'.  y  =  x2  =  9  and  &x  =  .001,  the  side 
x  becomes  x  +  A#.  Now  area  of  square, 

x*  +  2  x     (A#)   +   A*2 

(  (x  +  A*)2  =  (3  +  .ooi)2  =  9  +  2  (3)   (.001)  +  .000001 
32       =9 


By  subtraction  ; 
Dividing  by     A: 

A;y=  2  (3)   (.001)  +  .000001 
2  (x)  (A*)  +      A*2 

v=  .001,  we  get    —^  =  2  (3)  +  .001. 

A# 

Now  if  A#  becomes  dx,  then  the  number  of  zeroes  before 
the  i  in  the  last  term  would  be  infinite  and  we  would  have 


Now  3  is  the  length  of  the  side  x,  which  is  as  we  see 
introduced  into  the  calculation  in  a  perfectly  general  way, 
as  is  also  the  factor  2.  Thus  if  x  =  8  and  A*  =  .0000  1 

then  —  -  =  2  (8)  +  .00001 

A# 

2   (X)   +   A* 

and  similarly  for  any  other  values  of  x  and  Ax.  Hence 
it  would  seem  that  we  might  write  for  the  differential 

quotient  the  general  value  —  =  2  x,  where  x  represents 

dx 

the  length  of  a  side  at  any  moment.  If  x  =  7  then  2  x  = 
14,  and  since  dy  —  2  xdx,  we  find  that  the  rate  of  in- 
crease of  the  square  in  square  units  =14  times  the  rate 
of  increase  of  the  side  in  linear  units  at  the  moment  when 
the  side  is  7  units  in  length.  We  will  now  approach 


180  Elementary  Calculus. 

this  matter  more  generally   and   see  if  the  result   above 

indicated  is  a  rigid  truth. 

ART.  6.     Geometrical  view  of  the   differential  coefficient 

ofy=  x2. 
Suppose  we  have  a  square  the  side  of  which  is  x    (see 

Fig.  3).     The  area  x2,  we  call  y,  thus  we  have  y=  x2. 

Now  let  x  receive  an  incre- 
ment Ax,  then  x  +  Ax  can  be 
considered  as  the  side  of  a  lar- 
ger square  (x  +  A:*;)2.  Com- 
pleting the  construction  shown 
in  Fig.  3,  we  notice  that  the 
difference  between  the  squares 
(x  +  Ax)2  and  x2,  which  is 
(x  +  Ax)2  —  x2,  is  made  up 
Aa,  of  two  "rectangles  Pj  and  P2 
Fig.  3.  together  with  the  small  square 

S.     The  rectangles  have  each 

an   area  of  x .  Ax  and  the  square  S   of  Ax .  Ax  =  A.T2. 

These   parts   taken  together  represent  the  increase  Ay  of 

the  square  y  when  x  changes    to  x  +  Ax,  in  virtue  of  its 

increment  Ax.     We  thus  get  : 

Ay  =  2  .  x.  Ax-{-  Ax2 

(Increase  of  square  y)  —  (Two  rectangles  P1  P2.)  + 

(Square  S.). 

We  further  notice  that  the  square  S  is  much  less  in  area 
than  the  two  rectangles  Pt  and  P,.  Now  the  smaller  the 
increment  Ax,  the  narrower  become  the  rectangles  and 
the  less  the  relative  area  of  S.  This  is  easily  seen,  for  sup- 
pose Ax  is  exceedingly  small,  then  the  rectangles  Px  and 
Po  may  be  represented  by  long  thin  lines  (see  black  line 
Fig.  4),  while  S  is  reduced  to  their  intersection. 


Elementary  Calculus. 


181 


PL- 


If  now  we  consider  the  lines  representing  these  rectangles 
to  be  infinitely  thin,  then  the  sides  of  the  squares  become 
infinitely  short,  while  the  lines 
representing  the  rectangles  re- 
main of  finite  length,  hence  it 
would  take  an  infinite  number 
of  such  squares  to  make  one  of 

the    rectangles.       Clearly    the  PI 

square  S  tends  to  vanish  if 
the  rectangles  become  infinitely 
narrow,  that  is  if  Ax  changes 
to  dx  then  (dx)2  is  evanes-  pig.  4. 

cent,  that  is,  tends  to  "vanish. 

We  had  above,  Ay  =  2  xAx  +  (Ax)2. 

If  Ax  becomes  dx  then     dy  =  2  xdx 


< x 


and 


dx 


=    2  X. 


We  thus  find  that  if  y  =  x2,  then 


dx 


2  x.     In  other 


words  we  have  found  that  if  a  quantity  y  (in  our  case  the 
area  of  a  square)  is  dependent  upon  another  x  (here  the 
side  of  a  square),  in  such  a  manner  that  y  =  x2,  then  the 
rate  of  increase  of  y  at  any  moment,  compared  to  the  rate  of 
increase  of  x  at  the  same  moment,  is  =2  x,  which  latter 
quantity  is  called  the  differential  quotient  of  the  expres- 
sion y  =  x2,  or  more  generally,  the  differential  coefficient 
of  x2  with  respect  to  x. 

ART.  7.  Differential  coefficient  of  y  =  x2.  Analytical 
method. 

We  will  now  examine  a  general  analytical  method  of 
obtaining  the  differential  coefficient  of  x2  with  respect  to  x  in 
the  case  of  the  function  y  =  x2. 


182  Elementary  Calculus. 

Given  y  =  x2, 

then    y  +  A;y  =  (#  -f-  A#)2  =  x2  +  2 

now  y  +  A;y  =  x2  +  2 

and  =  x2. 


Subtracting;  Ay  ?=  2 

.-.    £*-  2*  +  A*. 

A# 

If  A#  becomes  dx  then  the  value  of  A#  alone  tends  to 
vanish  or  is  evanescent. 


Hence  again  we  find  if  y  =  x2,  then  the  differential 
quotient  of  the  expression  y  =  x2  is  2  x;  which  is  also  the 
differential  coefficient  of  x2  with  respect  to  x,  for  2  x  is  the 
multiplier  of  the  differential  dx  of  the  independent  variable 

x  when  we  write    -^-    =  2  ^  in  the  form  of  d-y  =  2  #  .  dx. 
dx 

ART.  8.     Differential  coefficient  oj  y  =  Xs. 

We  will  now  take  another  case;  if  y  —  j  (x)  and  the 
function  be  such  that  y  —  x3,  what  is  the  relation  of  dy 
to  dart 

Suppose  x  to  be  a  straight  line,  then  x3  will  represent 
the  volume  of  a  cube  =  y. 

Now  let  x  increase  by  A#,  then  x  +  A#  will  form  the  side 
of  a  second  larger  cube  whose  volume  is  y  +  Av. 

Now  if  we  examine  Fig.  5,  we  see  that  A;y  which  is  the 
difference  in  volume  of  the  two  cubes,  (x  +  A^;)3  and  x3, 
is  made  up  of  three  slabs  each  of  dimensions  x.  x  .  A:*; 
=  x2Ax  together  with  three  parallelopipidons  of  dimen- 
sions x  .  Ax  .  A^  =  x  .  A#2  and  of  one  cube  of  volume 
A#  .  Ax  .  A#  =  A^e3. 


Elementary  Calculus.  183 

Hence  we  have  \y  =  3  x2  Ax  +  3  x  A#2  + 

and  I=2=x2         x&x 


Fig.  5. 


If 


becomes  dx  then, 
dy 


dx 


=  3  *2  +  3  x  .  dx  +  (dx)2. 


Now  both  $x.dx  and  (dx)2  are  evanescent,  but  remember- 
ing the  ratio  of  the  infinitely  small  quantities  dy,  dx,  is  finite, 
it  is  in  fact  the  quotient  3  x2. 

Hence  if  y  =  x3  then  —  =  3  x2, 
dx 

or  dy  =  3  x2  dx. 

Therefore  the  differential  coefficient  of  y  =  x9,  with 
regard  to  x,  is  3  x2  and  the  expression  dy  =  3  x2dx  means 
that  at  any  moment  the  rate  of  increase  of  the  volume  in 


184  Elementary  Calculus. 

units  of  volume  is  3  x2  times  the  rate  of  increase  of  the  side 
in  linear  units. 

If  the  sides  be  2  inches  and  the  increment  A#  is  .001 

then  —2  =  3  x2  +  3  x  Ax  +  A^;2. 


=    12    +  .OO6   +  .OOOOOI. 

Obviously  if  Ax  becomes  evanescent,  the  value  of  the 
right  hand  member  becomes  =12. 

.'.    when  — -2   becomes  -^-,      then  -2-  =  12. 
A#  dx  dx 

This  result  we  could  obtain  at  once  from  the  previous 

expression      -*—  =  3  x2;  for  putting  x  =  2, 
dx 

we  get  |2L  =  3  (4)  =  12. 

Meaning,  that  at  the  moment  when  the  side  x  is  two 
units  in  length,  the  volume  of  the  cube  increases  12  times 
as  fast  in  units  of  volume  as  the  side  in  linear  units. 

ART.  9.  d.c.  of  y  =  x3,  analytically.  Orders  of  Infini- 
tesimals. 

If  y=**, 

then  y  +  Ay  =  (x  +  A^)3. 

.*.    y  +  Ay  =  y?  +  3 

y  =  Xs. 

Subtracting;      Ay  =  3  #2A#  +  3  # 
And  if  A.v  becomes  dx, 
then  Jy=3«8(Z»  +  3^  (^)2  +  (dx)9. 


Elementary.  Calculus.  185 

Now  dx  is  an  infinitesimal,  and  when  it  occurs  in  the 
first  power,  is  said  to  be  of  the  first  order;  similarly  (dx)2 
and  (dx)3  are  of  the  second  and  third  orders  respectively. 

Obviously  the  same  reasoning  that  causes  us  to  consider 
an  infinitesimal  of  the  first  order  as  unimportant  when 
compared  to  a  finite  quantity,  leads  us  to  regard  an  infin- 
itesimal of  any  higher  order  as  evanescent  when  com- 
pared with  one  of  lower  order.  Then  the  quantities 
3  x(dx)2  and  (dx)3  are  unimportant  terms  in  the  expres- 
sion 

dy  =  3  x2dx  +  3  x  (dx)2  +  (dx)3. 

Hence  dy  =  3  x2dx 

and  ^=3*2 

dx 

ART.  10.     The  d.c.  and  the  gradient. 

In  engineering  work  grades  are  often  described  by  refer- 
ring the  rise  in  level  of  a  point  to  its  corresponding  hori- 
zontal distance  from  some  fixed  position.  We  thus  speak 
of  a  grade  of  20  ft.  in  100  ft.,  meaning  the  slope  resulting 
from  arise  of  20  ft.  in  100  ft.,  or  i  ft.  in  5  ft.,  as  indicated 
in  Fig.  6,  and  measured  by  the  tangent  Z.  BAG.  The 


term  "  gradient  "  is  applied  to  the  numerical  value  of  the 
rati0;  vertical  rise         _    BC          (See  Rg.  6-) 

honzontal  distance          AB 

Now    tangent    BAG  =  -5£-  =   -  =  0.2,  and  since  the 
Ar>          5 

natural  tangent  of   (11°   19')  =  0.2   unit,    therefore,   the 


[86 


Elementary  Calculus. 


gradient  of  the  slope  AC  is  0.2,  and  the  angle  BAG  is  approx- 
imately 11°  19'. 

Suppose  a  straight  line  AB  to  make  an  angle  DCB  with 
the#-axis.     (See  Fig.  7.) 


Fig.  7' 


Let  the  co-ordinates  of  any  point  Q  on  AB  be  x  and  y. 
Let  x  he  increased  by  A#,  and  y  by  A;y. 

Completing  the  construction  shown  in  Fig.  7,  we  have 

tan  Z  DCB  =      ^-        -  (by  similar  triangles), 


and 


'         Ay  . 

r^ ^—  • 


QR 


Hence  =%  =  tangent  Z  DCB. 

If  the  increment  A#  becomes  infinitely  small,  then 
&.  =  tangent  ZDCB. 

This  means  that  in  the  case  of  a  linear  function,  that  is, 

a  function  whose  graph  is  a  straight  line,  the  ratio  of  an 

'  infinitely  small  increment  of  the  y-ordinate  to  dx  gives  the 


Elementary  Calculus. 


tangent  of  the  angle  which  the  straight  line  makes  with  the 
#-axis,  and  therefore  its  gradient. 

We  will  now  test  this  numerically  by  the  following 
example. 

Given  the  linear  function,  y  =  0.7  x  +  2,  to  find  the 
differential  coefficient  with  respect  to  x,  namely,  the  value 


of    2Z-,    and 

dx 

hence  the  gradient  of  the  line. 

We  have 

y=  Q.JX  +  2, 

then 

y  +  Ay  =  0.7  (x  +  A#)  +  2. 

Hence 

y  +  Ay  =  0.7  x  +  0.7  A#  +  : 

But 

y  =  0.7*  +  2. 

Subtracting; 

Ay  =  0.7  A#. 

•'  fir0-7- 

and 

£  =  o.7. 

2. 


o    1 


"A  -2    -l 


Now   0.7   is   the   approximate   natural   tangent   of  35°. 
Hence  by  differentiating  the 
function  y  =  0.7  x  +  2   we 
have   not    only    found    the 
ratio  of  the  increase  of  the 
ordinate    to    the     abscissa 
at  any  moment,  but  also  the 
gradient    of    the    line    and    . 
hence  the   angle  it  makes 
with  the  #-axis. 

The  line  AB,  Fig.  8,  was 
plotted  from  the  equation 
y  =  0.7  x  +  2,   and  the  angle  BA#  will    be   found,   upon 
measurement  with  a  protractor,  to  be  approximately  35°. 


Fig.  8. 


1 88  Elementary  Calculus. 

ART.  ii.     The  gradient  of  a  curve. 

Suppose  we  have  two  bodies,  "Bl  and  B2,  travelling  in 
parallel  paths,  the  former  with  an  accelerated  velocity  of  2  ft. 
per  second  per  second  and  the  latter  with  a  uniform  velocity 
of  2  ft.  per  second.  Further,  imagine  that  Bx  starts  upon  a 
line  AjA2  (see  Fig.  9),  while  B2  starts  one  foot  to  the  left 
of  it  but  at  the  same  moment. 


B2 


Fig.  9. 

In  the  first  case,  that  of  B  v  where  the  velocity  is  acceler- 
ated, we  have  5  =  \  at2,  where  a  =  2  is  the  acceleration, 
hence  s  =  J  (2 )  /2,  and  therefore,  s  =  /2. 

In  the  second  case,  the  velocity  is  constant,  and  we  have 
the  space  traversed  by  B2  expressed  by  the  equation  s  =  vt, 
and  since  v  =  2,  we  have  s  =  2  t. 

The  following  table  gives  the  spaces  traversed  by  B, 
and  B2  at  the  conclusion  of  different  time  intervals. 

Br  B2. 

Space  traversed  from  Space  traversed  from 

rest  at  the  end  of  rest  at  the  end  of 

J  second   =  J  ft.  J  second   =  i  ft. 

1  second  =  i  ft.  i  second   =  2  ft. 

2  seconds  =  4  ft.  2  seconds  =  4  ft. 

3  seconds  =  9  ft.  3  seconds  =  6  ft. 

In  Fig.  9,  we  have  depicted  the  relative  positions  of  the 
two  bodies  E1  and  B2  graphically,  showing  a  portion  of  their 
paths,  and  using  the  data  given  in  the  above  table.  Notice 


Elementary  Calculus. 


189 


that  during  the  first  second,  Bt  travels  slower  than  B2,  and 
that  B2  has  caught  up  with  Bt  at  the  end  of  the  first  sec- 
ond, and  for  one  instant  of  time  the  two  are  abreast,  and 
travelling  with  the  same  velocity,  after  which  the  speed  of 
Bj  is  greater  than  that  of  B2  and  is  constantly  growing,  as 
shown  by  the  increasing  distance  covered  in  each  ensuing 
second. 


SIN  FEET 


7/ 

/R 


B2/H 


7 


Fig.  10. 


Plotting  the  values  given  for  s  and  /  in  the  above  table 
we  obtain  in  the  case  of  Bt  a  curve  (see  Fig.  10),  and  in 
that  of  B2  a  straight  line;  this  latter,  it  will  be  noticed, 
touches  the  curve  at  the  point  P;  which  point  corresponds 
to  the  positions  of  the  two  bodies  when  they  are,  for  an 
instant  of  time,  one  foot  from  the  line  A,A2  and  traveling 
with  the  same  velocity. 


190  Elementary  Calculus. 

We  have  already  said  (Art.  10)  that  the  gradient  of  a 
line  is  measured  by  the  tangent  of  the  angle  that  the  line 
makes  with  the  abscissa;  but  if  a  line  is  a  geometrical  tan- 
gent to  a  curve,  then  at  the  point  of  tangency  the  two  have 
the  same  direction.  Hence  the  slope  of  the  geometrical 
tangent  to  a  curve,  at  a  point,  shows  the  steepness  of  the 
curve  at  that  point,  but  the  gradient  of  the  line  is  measured 
by  the  tangent  of  its  abscissa  angle.  We  thus  have  the  fol- 
lowing definition:  The  gradient  oj  a  curve  at  any  point 
is  measured  by  the  tangent  oj  the  angle  which  the  geometrical 
tangent,  at  thai  point,  makes  with  the  abscissa. 

Now    the    gradient    of    the    line    NH    is   measured    by 

tan    MNP  =        : —  =  —  =  2,  and  this  quantity  is  also  a 

NM         % 

measure  of  the  gradient  of  the  curve  at  the  point  P,  from 
the  above  definition. 

Let  us  now  take  increments  to  the  ordinates  of  P;  let  the 
time  increment  of  /be  A/  =  PQ,  in  both  the  case  of  the 
curve,  and  that  of  the.  line;  for  the  space  increment  we 
have,  for  the  line,  As  =  QR,  and  for  the  curve,  As  =  QK. 

Hence  for  the  line,    — -  =  -£ —  , 
A/         PQ 

As        QK        QR  +  RK 
for  the  curve,  ==  ~~ 


Now  clearly  in  this  case  if  A/  is  infinitely  small,  then 
the  latter  expression  becomes  —  ,  as  can  be  inferred  from 
the  figure. 

Hence  —  at  the  point  P  has  the  same  value  for  both  the 

line  and  curve,  namely  -—  =  2. 

at 


Elementary  Calculus. 


191 


That   is,  the    value   of    the  differential  quotient   of  the 


function  s  =t2,  for  the  point  P  (i,  i),  namely  — 

dt 


2,  is 


the  tangent  of  the  angle  the  geometric  tangent  makes 
at  P. 

We  will  now  see  if  this  statement  is  susceptible  of  a 
general  application. 

Let  y  =  /  (x)  be  any  curve  of  which  a  portion  of  the 


Fig.  ii. 


graph  is  shown  in  Fig.   n.     Suppose  the  point  P  upon 
y  =  }  (x)  has  the  co-ordinates  OM  =  x  and  MP  =  y. 

If  MB  =  A*  then  QK  =  Ay,  and  the  ratio  of  the  rate 
of  increase  of  the  function  y  to  the  rate  of  increase  of  the 

independent  variable  x,  will  be  expressed  by   — ^  •    Now 

Ax 

Ay 

-— ^-  =  tan  KNB ;  which  latter  is  the  tangent  of  the  angle 

that  the  geometrical  secant  NK  makes  with  the  #-axis. 


192  Elementary  Calculus. 

The  value  of  —  •—  will  depend  upon  the  size  of  the  incre- 

A* 

ment  A#,  as  we  have  already  seen,  except  in  the  case  of  a 
straight  line  when  the  function  is  linear.  Further  the 

value  of    --2-  is  dependent  upon  the  position  of  the  point  P. 

Ax 

as  can  be  readily  inferred  from  the  figure,  for  if  P  were 
moved  to  the  right,  then  an  increment  A#  would  bring 
about  an  immensely  increased  corresponding  increment, 
Ay,  because  of  the  steeper  slope  of  the  curve,  and  there- 


fore —-   would  assume  a  greater  value. 

A# 

If,  however,  A#  is  gradually  decreased,  then  the  point  K 
will  continually  approach  the  point  P,  while  the  secant 
NK  will  cut  the  abscissa  at  a  more  and  more  acute 
angle,  until  finally,  when  A#  =  dx,  the  secant  will 
take  its  limiting  position  AH,  which  is  the  geometric 
tangent  to  the  curve  y  =  /  (x)  at  the  point  P,  and  we  have 


It  is  important  to  notice   that  the  value  of  -^  depends 

dx 

wholly  on  the  direction  of  the  curve  at  the  point  P,  and, 
therefore,  expresses  its  gradient  at  this  point. 

Hence,  if  y  =  j  (x),  then  the  differential  coefficient  of 
this  junction  is  equal  to  the  tangent  of  the  angle  which  the 
geometric  tangent  to  the  curve  at  any  point  upon  it  makes 
with  the  x-axis,  while,  at  the  same  time  it  expresses  the 
gradient  of  the  curve  at  that  point. 

From  Art.  9,  we  know  that  if  y  =  x3  then  -*    =  3  x2; 

(tOC 

putting  x  =»  i.i   we  find  3  x2  =  3  (i.i)2=  3.63,  therefore 


Elementary  Calculus.  193 

—  =  3-63;  which  on  referring  to  a  table  is  found  to  be  the 

doc 

natural  tangent  of  74°  36'. 

We  thus  have  found  that  given  y  =  x?,  the  ratio  of 
the  rate  of  increase  of  the  ordinate  to  that  of  the 
abscissa  at  a  point  where  abscissa  is  i.i,  is  3.63.  This 
latter  is  the  gradient  of  the  curve  at  that  point,  while 
the  geometrical  tangent  makes  an  angle  of  74°  36'  with 
the  ^-axis. 

Let  us  test  the  above  calculation  by  actually  plotting  the 
curve  and  drawing  the  tangent.  Fig.  12  shows  a  part  of 


Fig.  12. 

the  curve,  while  P  is  that  point  whose  abscissa  is  i.i.  If 
the  angle  KRx  be  measured,  it  will  be  found  to  be  about 
20°,  but  the  angle  which  the  tangent  to  the  curve  at  P 
makes  with  the  re-axis,  is,  according  to  our  previous  calcu- 
lation, 74°  36';  the  discrepancy  is  due  to  the  fact  that  the 
unit  of  measurement  used  on  the  #-axis  is  10  times  that 
used  on  the  y-axis. 

In  order  that  the  tangent  should  represent  the  true 
gradient  of  the  curve  at  P,  we  must  refer  the  ordinates  and 
abscissas  to  the  same  scale,  or  we  will  not  obtain  the  true 
comparative  rate  of  increase  of  y  to  x.  Tan  20°  =  0.363 
(nearly),  or  -^  of  the  true  value. 


194 


Elementary  Calculus. 


In  order  to  make  this  important  point  quite  clear,  we 
have  plotted  the  curve  y  =  x?  a  second  time  (see  Fig.  13), 


!R 


Fig.  13. 

and  have  used  the  same  scale  for  both  ordinates  and 
abscissas.  Upon  measuring  the  angle  PR#  with  a  pro- 
tractor it  will  be  found  to  be  74°  36'  approximately,  which 

corresponds  with  the  result    —  =  3.63. 


ILLUSTRATIVE    EXAMPLES. 

I.   Derive  the  differential  coefficient  of  the  function 

y=  2  x2  —  $x  +  i. 
Now,  y  +  Ay  =  2  (x  +  Ax)2  —  3  (x  +  A#)  +  i. 

/.    y  +  Ay  =  2  #2  +  4  #  A#  +  2  A:v    —  3  #  —  3  A#  + 1 
but,  y  =  2x2  —  3X  +1 


Elementary  Calculus.  195 

Subtracting;       Ay  =  4  x  A#  —  3  A#  +  2  A#2. 

.-.|2  =  4*-3+*A*. 
If  A#  becomes  <fo,  then  2  <fo  is  evanescent. 
Hence  — -^  =  4  #  —  3. 

f/JC 

II.  Find  the  gradient  of  the  curve  x2  —  x  +  2  =  y  at 
the  point  where  x=  1.15,  and  the  angle  the  geometrical 
tangent  at  this  point  makes  with  the  #-axis. 

y  =  ^2  -  X  -f  2j 

y  +  Ay  =  O  +  A*)2  -  (x  +  A*)  +  2, 

y  +  Ay  =  x2  +  2  #  A#  +  A,T2  —  ^c  —  A^  +  2, 

>>  =   X2 —  .V 4-  2. 

.'.  Ay  =  2  a;  A#  —  AJC  +  A^w. 

-^  =  2^  —  1+  A.T.     Hence  -^-  =  2  ^  —  i. 

A^  ^ 

To  find  the  gradient  of  the  curve  at  the  point  where 
x  =  1.15  we  substitute  as  follows: 

-2-  =  2  x  —  1=2  (1.15)  —  i  =  1.30. 

Hence  1.30  is  the  gradient  required,  and  since  tan  52° 
26'  =  1.30,  we  find,  therefore,  that  the  geometrical  tan- 
gent at  the  point  where  x=  1.15  makes  an  angle  of  52° 
26'  with  the  #-axis. 

III.  Find  the  rate  at  which  the  area  of  a  square  is  in- 
creasing at  the  instant  when  the  side  is  6  feet  long,  suppos- 
ing the  latter  to  be  subject  to  uniform  increase  of  length  at 
the  rate  of  4.5  feet  per  second. 


196  Elementary  Calculus. 

Let  %  =  length  of  side, 


y  =  x2  =  area. 
By  Art.  7,  dy  =  2  x  dx, 

that  is,  the  rate  of  variation  of  area  =  2  x  times  the  rate  of 
variation  of  the  side. 

Substituting  the  given  values,  we  get 

dy  =  2  (6)  (4.5)  =  54  Sq.  ft.  per  second. 

EXERCISE  I. 

Find   the   differential   coefficient   of   the   following   five 
functions  by  the  method  of  Art.  7. 

1.  y  =  2  x2  -  3. 

2.  y=    (x-    2)     (A?  +  3). 


#  —    I 

-   — 

X  +  I 

6.    Plot  the  graph  of  x2  +  3  x  —  2  =  y. 

(a)  What  can  you  tell  about  the  roots  of  the  equation 
from  the  appearance  of  the  graph  ? 

(b)  Find  the  general  expression  for  the  gradient  of  the 
curve  at  any  point. 

(c)  Find  the  angle  which  the  geometrical  tangent  makes 
with  the  curve  at  those  points  on  it  where  x  =  o,  x  =  —  }, 

X  =    -    2,   X  =    ~    2- 

(d)  Draw  tangents  at  the  points  where  x  =  —  f  and 
x  =  —  2,  and  test  your  answers  to  question  c  by  actual 
measurement. 

(e)  What  effect  would  it  have  upon  the  gradient  of  the 
graph  at  ^  any  point,  if  the  scale  for  the  ^-axis  was  made 
10  times  as  large  as  that  of  the  ^-axis  ? 


Elementary  Calculus.  197 

(/)  If  y  —  f  (x)  and--  -   =  a  for  a  certain  x  value,  what 
ax 

does  this  imply? 

7.  Differentiate  the  function  s  =  J  at2  with  respect  to  t. 
What  does  the  result  mean  ? 

8.  A  man  cuts  a  circular  plate  of  brass  the  diameter  of 
which  is  4  inches;  after  heating  he  finds  the  diameter  to 
have  increased  by  .006  of  an  inch.     What  is  the  increase  of 
area? 

9.  If  x  be  the  side  of  a  cube  which  is  increasing  uni- 
formly at  the  rate  of  0.5  inch  per  second  per  second,  at 
what  rate  is  the  volume  increasing  at  that  instant  when 
the  side  is  exactly  2  inches  in  length? 

10.  If  a  body  travels  with  an  accelerated  velocity  of  2 
ft.  per  second  per  second,  and  we  call  the  space  traversed 
at  the    end  of  the    first   second  s,    show  by  arithmetical 
computation  that  if  As  is  any  positive  increase  of  s,  then 

As 

—  approaches  more  nearly  the  actual  momentary  velocity 

of  the  body  at  the  end  of  the  first  second,  the  smaller  As 
is  taken. 


CHAPTER  II. 
DIFFERENTIATION. 

I.    Algebraic  and  Transcendental  Functions. 

ART.  12.  An  Algebraic  Function  is  one  in  which  the 
only  operations  indicated  are,  addition,  subtraction,  multi- 
plication, division,  involution,  and  evolution;  further,  such 
a  function  must  be  expressed  by  a  finite  number  of  terms, 
and  any  exponents  involved  must  be  constant.  Examples 
of  algebraic  functions  are, 

fy ' y%     _|_     -7      ft    j 

x2  +  2  x,    (x  —  m}*.    (x  —  n}$.      — ^ . 

(*-4) 

In  distinction  to  the  above  we  have  the  so-called  Tran- 
scendental Functions,  which  cannot  be  expressed  algebrai- 
cally in  a  finite  number  of  terms;  examples  of  which  are  as 
follows: 

sin  x,  tan  x,  vers  x,  loge  x,  ex. 

The  Binomial  Theorem. 

In  works  on  algebra  a  general  proof  of  the  following 
expansion  may  be  found: 

(a  +  bY  =  an  +  nan~l  b  +  n  (n  ~  *).  an~2b2 

I  .  2 


1.2.3 

For  convenience  we  will  put  n  =  Cv  —          —  =  C2,  etc. ; 

i  .  2 

we  thus  get, 

(a  +  b)n  =  an  +  C1  a"-1  b  +  C2  an~2  b2  +  C3  an~3  &»  +  .., 
198 


Elementary  Calculus.  199 

ART.  13.    Differentiation  of  axn  and  xn. 
If  y  =  axn, 

then  y  +  Ay  =  a  (x  +  Ax)n. 

Expanding   the  right-hand  "member,   as  explained  in  the 
previous  paragraph,  and  multiplying  through  by  a,  we  get 

y  +  Ay  =  axn  +  a  Q  x"-1  A  x  +  a  C2  xn~2  (A  x)2 

+  aC3  xn^  (A*)3  .+   .  .   . 

But,  y  =  ax\  _ 
.-.  A)/  =  a  Qx"-1  Ax  +  a  C2xn-2  (Ax)2 

+  a  C3xn^  (Ax)3+  .   .  . 
and  =  a      xn~l  +aC  xn~2kx  +  a  C 


If  Ax  becomes  dx,  then  all  the  terms  of  the  right-hand 
member  after  the  first  are  evanescent  (Art.  6);  and  remem- 
bering Cj  =  n  (see  Art.  12),  we  get 


dx 

Now  if  in  the  function  y  =  axn,     a  =  i, 
we  get  •      y  =  xn, 

and  $.=  nx»-i. 

dx 

To  differentiate  y  =  xn  with  respect  to  x.  First,  multiply 
x  by  the  index  and  then  obtain  the  new  power  by  diminish- 
ing the  index  by  unity. 

Example  :  y  =  x4  ;   -^-  =4  x4-1  —  4  x3. 
rfx 

To  differentiate  y  =  ax";  differentiate  the  function  xn 
multiply  the  result  by  the  constant. 

Example:  y  =  5  x3  ;       -  =  5  (3)  x3-1  =  15  x2. 


2oo  Elementary  Calculus. 

The  results  above  obtained  are  true  for  all  values  of  n, 
whether  positive,  negative,  or  fractional  ;  the  proof  of  the 
latter  two  cases  is  simple,  and  is  left  as  an  exercise  for  the 
student. 

Examples  :y=  -  x-3:  &.=    -•!  x-3-1  =  -  ^  x~* 


2 

y=  —  xr 
Example:*  y  =  2  V  ot 

*  .  *L  _  9 
'  <fo       5 

;2    .'.  y=  2X- 

2              q  ., 
.      ^       3C 

I;^L     =     i 

'  ^         3 

6 

5 

nt 

ART.  14.     Differentiation  of  a  constant. 
We  have  denned  a  constant  as  a  quantity  which  does  not 
change  or  alter  its  value.     Hence  if  k  is  a  constant,  A& 

A    Ak  ,,       c        dk 

=  o  and    -  —  =  o,     therefore  —  =  o. 

A#  dx 

ART.  15.     Differentiation  of  a  sum. 

Suppose  y  =  u  +  v,  when  both  u  and  v  are  functions 
of  x.  Now  if  x  becomes  x  +  AJC,  then  w  and  v  become 
u  +  AM  and  v  +  Av,  respectively,  and  we  get, 


But  =  z*  +  v. 


/.  Ay  =  -I./  +  Az'. 

Divide  bvA^;     /.    ^=^+^. 
A^      Ax      Ax 

*  If  the  function  involves  a  radical  which  can  be  reduced  to  the 
u 

form  xv  ,  then  express  the  radical  as  a  fractional  power  and  proceed 
as  above. 


Elementary  Calculus. 


201 


If 
then 


becomes  dx, 


dy         du    ,   dv 
dx         dx       dx 
In  a  similar  manner  we  can  show  that  if 

y = u±v ±w± ... 


then 


dy    _  du     .    dv     .    dw     . 
dx         dx      "   dx    '  ~  dx 


Hence,  the  differential  coefficient  of  the  sum  of  several 
functions  is  the  sum  of  the  differential  coefficients  of  the 
several  parts,  due  regard  being  given  to  the  signs. 

Example:  y  =  3  x3  —  5  x2  +  2  x  +  3. 


By  Art.  14, 


dx 


dx 


=0. 


=  9  x2  — 10  x  +  2. 


ART.  1  6.     Differentiation  of  a  product. 

If  y  =  u  .  v  where  u  and  v  are  each  functions  of  x,    re- 


quired the  value  of 


dy_ 
dx 


In  order  to  obtain  a  clear  idea  of  the  meaning  of  the 
above   function,   suppose  u  =  5  x   and  v  =  3  x.     Then 
G  AMF 

Ar 


x  =  5v  a*" 

Fig.  14. 

u  .  v  can  be  geometrically  represented  by  a  rectangle  ABCD 
(see  Fig.    14),  two  of  whose  opposite  sides  are  each  of 


202  Elementary  Calculus. 

length  u  =  5  x,  while  those  adjacent  are  represented  by 
v=  3*. 

If  x  is  increased  by  Ax  then, 

u  +  Aw  =  5  (x  +  A#)  =  5  #  +  5  A#  =  AE, 
and  v  +  kv  =  3  (x  +  A#)  =  3  x  +  3  A#  =  AG. 

Hence  Aw  =  5  A#  and  Ai;  =  3  A#. 

Completing  the  figure  as  shown,  we  see  that  A;y,  which  is 
the  difference  in  area  between  the  rectangles  AEFG  and 
ABCD,  is  made  up  of  three  small  rectangles  whose  areas 
are  obviously  3  x  (5  A#),  5  x  (3  A#),  and  (5  A#)(3  A#), 
respectively. 

Hence  Ay  =   3  x  (5  A#)  +  5  x  (3  A*)  +  (5  A#)  (3  A*). 
.'.f^=   3*  (5)  +5*  (3)  +5  (3  A*). 

Now  if  A#  is  a  small  decimal  say  o.ooooooi,  clearly  the 
last  term,  which  represents  the  least  of  the  rectangles,  will 
tend  to  vanish;  therefore,  if  A#  becomes  dx,  we  have 

^  =  3*  (5)  +5*  (3)       ....     (i) 
But  u  =  5  x  and  v  =  3  x, 

and  the  differential  of  the  first  function  is   —    =  5  and  that 

dx 

of  the  second  is  —  =  3. 
dx 

Hence  substituting  in  (i); 

^_  =  v  .  ^L  +  u  .  —  - 
<fo;  cfo  d# 

In  general  if         y  =  u  .  v  ; 

y  +  Ay  =  (w  +  Aw)  (v  +  Av). 


Elementary  Calculus.  203 

/.  y  +  A?  =  uv  -f  vAw  +  wAv  +  Aw.  Av, 


but 


Hence  A;y  =  vAw  +  u&v  +  Aw  .  Av. 

Dividing  by  A#; 


If  A#  becomes  doc  then  —  Av  =  —  dv  which  is  evanes- 

A#  dx 

cent,  for  although  the  quotient  —  is  finite,  it  is  multiplied 

dx 

by  the  differential  dv,  and  therefore  tends  to  vanish. 

TT  dy          du  dv 

Hence  -^-  =  v  —  •    +  u  -- 

dx         dx  dx 

Again  if  y  =  u  .  v  .  w\ 

then  putting       u  .  v  =  z 
we  get  y=z.  w, 

dy  dz          dw  ,  x 

and  -f-  =  w  —  --  h  z  —        .....     (a) 

dx  dx          dx 

But  since  z=  u  .  v, 


.  . 

dx          dx  dx 

Substituting  this  value  of   —   in  (a) 
dx 

dy  du  dv  dw 

we  get,       _^_  =  vw  —  +  uw  —  •  +  uv  —  •  . 
dx  dx  dx  dx 

A  like  form  can  be  found  for  the  differential  coefficient  of 
any  number  of  variables. 

Hence,  the  Differential  Coefficient  of  a  Product  of  several 
variables,  is  the  sum  of  the  products  of  the  differential  coeffi- 
cients of  each  variable  multiplied  by  all  the  others. 


2O4  Elementary  Calculus. 

Example:          y  =  (3  x  +  2)  (5  x  —  6) 
Z=  (5* -6)  +  3)    +  (3*  +  ,) 


=  (5* -<*)  (3) +  (3*  +  *)  (s). 

dy 

.'.  -^~  =  30^  —  8. 

ART.  17.   Differentiation  of  a  quotient. 
Let  y  =   - , 

when  z/  and  v  are  functions  of  #. 
We  have,        u  =  vy, 

dx  '  dx  '  dx 

dx        dx  dx 

but  ?  =  ->     /.z;.^L=  ^--£   .   ^L, 

v  '  d#        dx       v    '   dx 

du_  _  u_  dv 

dy        dx       v   dx 

and  -J—  — 

dx  v 

Multiplying  numerator  and  denominator  by  v  we  get 

du  dv 

v  —  —  u  — 
dy  dx  dx 

^  =      —~ 

Hence,  the  Differential  Coefficient  of  a  fraction  whose  num- 
erator and  denominator  are  variables,  is  equal  to  the  product 
of  the  denominator  and  the  differential  coefficient  of  the 
numerator  minus  the  numerator  times  the  differential  coeffi- 
cient of  the  denominator,  the  whole  divided  by  the  square 
of  the  denominator. 


Elementary  Calculus.  205 

:  c  is  a  com 
of  a  constant  is  zero,  we  get, 


If  y  =  -  where  c  is  a  constant,  then,  since  the  differential 


dx          c   dv 


Example:          y  = 


dx  v2  v2  dx 

i  —  x 


i  +  x2 


dy  _    _  dx  _  dx 

dx~~  (i  +  x2)2 

_  (i  +x2}  (-  i)-  (i-*)  (2*) 

(I    +  X2)2 

dy  _  x2  —  2  x  —  i 
dx  ~        (i  +  x2)2 

ART.  1  8.     Differentiation  of  a  function  of  a  function. 


Suppose  we  wish    to    evaluate        x2  +3^  +  2,    when 
x  =  1,2,  etc.      Putting 


V x2  +  3  x  +  2  =  y  and  x2  +  $x  +  2  =  2, 
then  y  =  "2/z 

if  x  =itz=  6       and  y  =  ^/6  =  1.817 

x  =  2,  z  =  12     and  y  =  v  12  =  2.289. 

Clearly  z  is  a  function  of  #,  and  further  the  value  of  y 
depends  upon  that  of  z,  hence  y  is  also  a  function  of  z.  We 
thus  see  that  y  is  a  function  of  z  which  in  turn  is  a  function 
of  x,  and  we  therefore  say  that  y  is  a  function  of  a 
function. 

This  latter  term  is  sometimes  puzzling  at  first,  and  care 


206  Elementary  Calculus. 

should   be   taken   that   it  is   thoroughly   understood.     Let 
us  take  the  general  case 

y  =  F  (z) 
and  z  =  /  (x). 

Now  if  x  undergoes  a  small  change  in  value  then  z  will 
change  likewise. 

If  x  becomes  x  +  A#, 

z  becomes  z  +  Az, 

r    ,  Ay  _  A)/     Az         [An     identity,    found 

A#       Az      A#  by  multiplying    and 

dividing  — ^-  by  Az.] 

Ia3£ 

and  if  Ax  becomes  dx, 

then  **-  =  &  .  &_. 

o*         dz       dbf 

Hence,  i/  y=  F(z)  and  z=  /(^),  the  differential  coeffi- 
cient of  y,  with  respect  to  x,  is  equal  to  the  product  of  the 
differential  coefficient  of  y  with  respect  to  z,  times  the  differ- 
ential coefficient  of  z  with  respect  to  x. 

Example  I :  y  =  \/u,          to  find  -2-  , 

dx 

where  x2  +  3  =  u. 

Since  y  =  \/u, 

we  have,  y  =  F  (u)  and  u  =  }(x). 

From  the  above,      $L  =  *L.   *L, 

dx       du      dx 

but  y  =  u*. 


Elementary  Calculus. 

and  since  u  =  x2  +3, 

/.  -^-  =  2  x. 


207 


'  dx       \A2  +  3 
In  general  we  would  proceed  thus: 
Given,  y  =  \/x*  +  3, 


Example  II:  y=  (x3  +  2)   (#  +  3)3. 
Here  we  have  a  product,  hence  by  Art.  16  we  get, 


(i) 


As  the  expression    (x  +  3)3  is  a  function  of  a  function, 
we  have, 

and 


dx 


Substituting  (2)  and  (3)  in  (i)  we  find, 

dx 
and   &-  =  6  x5  +  45  x*  +  io&  x*  +  87  x2  +  36  x  +  54. 


208 


Elementary  Calculus. 


EXERCISE  II. 

5  x?  +  3  x2  -x+2.     2.   y  =  ax2  +  bx  +  c. 


3  o    —  5  x   +  7  -  8  a*  +  2  #  -I. 


22. 

23. 
24. 


(3  X  —   2). 

y  =  x2(2  x3  +  i). 

y=  (x  +  i)   Oc2-  *  +  i). 


25.    ?  = 


Elementary  Calculus. 


209 


26.  y 

27.  y 

28.  y 

29.  y 

30.  y 

31-  y 
32.  y 

33-  y 

34-  ^ 

35-  7 
36.   y 

37-   y 

38.  y 

39-   y 
40.   y 


==   X2    \/2  X2  —   I. 
X2 

=    x2  —  3  x  +  i 

X2  -   I 

b  -  x 
'    b  +x' 


v/ 


b  -  x 

b  +x 
(x2-b}2 
(x*-b)* 


Vx  +  i 

X2 

\/x  —  i 

\/X  +  I 

X 


\/a2  +  x2  —  x 


v/ 


I  —  \/x 
i  +Vx 

v^ 


-x/i- 


^ 


\ 


+  Vi-* 


X\/ 


210 


Elementary  Calculus. 


II.        Differentiation  of  Transcendental  Functions. 

/THT         7        f  sin  a        •,  tana      , 

ART.  19.     The  value  of  -     -  and  -     -   when  a  becomes 
a.  a 

infinitely  small.  In  higher  mathematics,  angular  meas- 
urement is  always  expressed  in  radians.  The  choice  of 
the  radian  as  a  unit  possesses  many  advantages.  It  en- 
ables us,  for  example,  to  compare  directly  the  rate  of 
change  of  a  sine  with  the  rate  of  change  of  its  corre- 
sponding angle. 

It  is  important  that  the  student  should  now  examine  the 

values  of  the  two  expressions 


and    an  a    as  a  dimin- 


Fig.  15. 


But  coso  =  i,  hence 


a  a 

ishes. 

A  glance  at  Fig.  15  will  show  that 
for  any  angle  a, 

sin  a  <  a.  <  tan  a. 
Dividing  by  sin  a,  we  get 
sin  a;          a         sin   a      i 
sin  a      sin  a       cos  a  sin  a 
a  i 


i  < 


sin  a      cos  a 


cos  o 


=  i;  and  as  a  diminishes,  the 


more  nearly  does—      -  approach   the  value   i,  and  when 


cos  a 


a  is  infinitely  reduced,  --  =  i;  therefore;  we  may  put 


cos  a 


the  expression  --  or      -     =  i  when  the  angle  a.  is  infi- 
sma          a 

nitely  small,  for 


sin  a 


stands  constantly  between  i  and  a 


Elementary  Calculus.  211 

quantity, ,   which    continually    approaches    i,   as 

(cos  a) 

shown  by  the  inequality,  hence  — ^ —  must  itself  approach 

i  in  advance  of  — - — ,  and  will  reach  it  when ar- 

cos  a  cos  a 

rives  at  that  value. 

Again,    ^^L  =  J1BJL  .  _J: — y  but  we  have  seen  that 
a  a          cos  a 


each  of  the  expressions      -and  —  -  —  tends  to  approach 

a  cos  a 

the  value  unity  as  the  angle  diminishes;  hence  we  may  put 

-  =  i  when  a  is  infinitely  small. 
a 

ART.  20.     Differentiation  of  y  =  sin  x  and  y  =  cos  x. 
If  y=  sin  x, 

then          y  +  Ay  =  sin  (x  +  A#). 

y  +  Ay  =  sin  x  cos  A#  +  cos  x  sin  A#. 
And  y  =  sin  x. 

.*.  Ay  =  sin  #  cos  A^  —  sin  x  +  cos  x  sin  A#. 
.'.  Ay  =  sin  x  (cos  A#  —  i)  +  cos  x  sin  A#. 


Hence  -         =  (cos  A^  -  i)  + 

A#  AjC 

but  when  A#  is  infinitely  small, 


/.  when  Ax  becomes  dx,  then 

^L  =  i(0)  +cos*    ......     (i) 

dx 

.-.  -2- 
dx 


212  Elementary  Calculus. 

In  an  exactly  similar  manner  to  the  above  we  may  show 

that  if  y  =  cos  x,  -%-  =  —  sin  x. 

dx 

ART.  21.    Differentiation  of  y  =  tan  x  and  y=  cot  x. 
If  y=  tan  x, 


then 
By  Art.  17,     - 


COS  X 

cos  x  .  d  (sin  x}  —  sin  xd  (cos  x^ 


dy    _  cos  x  .  cos  x  —  sin  x  (—  sin  x) 
dx  cos2  x 

dy    __  cos2  x  +  sin2  x 
dx  cos2  # 

Jy         g?  (tan  #)  _       i  2 

</#  <fo  cos2  # 

In  like  manner,  if  y  =  cot  xt  we  may  show  that 

dy  i  . 

-r-  = — —  =  —  esc2  x. 

dx  sm2  ^ 

ART.  22.    Differentiation  of  y  =  sec  #  aw^  y=  cosecx. 
If  y  =  sec  ^,  then  y=- 

Differentiating,  we  find 

dy        sin  x 

-f-  =  -    - —  =  tan  x  sec  x. 

dx       cos2  x 

[Since  s-^  =    ***-  .  -1-  =  tan  x  sec  x.] 


Elementary  Calculus.  213 


Similarly,  when  y  =  cosec  x,  then  y  =  -^— 


and 


dx  sin2  x 

The  following  convenient  table  should  be  committed  to 


memory  :  * 


=  c     x    -2-  =  — 


y  =  sn  x]  --  =  cos  x       y  =  cos  x;  --  =  —  sn  x 
dx  dx 

y  =  tan  x;  -2-  =  sec2  x      y  =  cot  x;  -%-  =  —  Csc2  x 
dx  dx 

y  =  sec  x:  -*-.  =  tan  #  sec  # 


y  =  cosec  x:  --  =  —  cot  #  esc  #. 
dbc 

Since  vers  #  =  i  —  cos  x,  \i  y  =  vers  x, 

we  have  y  =  i  —  cos  x,  and,  therefore,  -2-  =    sin  x: 

dx 

also  if  y  =   covers  x=  i  —  sin  x,    —  =  —  cos  x. 

dx 

• 

EXERCISE  III. 

i.   y=  tan  (bx).  2.    y  =  cos  —  • 

x 

3.   y  =  sin  (3  #2).  4.   y  =  tan  \/W. 

5-  y  =  3  cos  (xH)-  6.   v  =  6  sin  —  • 

.r 


7.   y  =  sin  (i  +  ax2).       S.   y  =  cos  4  /—  « 

V  x 

g.   y  =  sin5  #.  10.   y  =  cos4  a^  .  x2. 

*  Note  that  the  differential  coefficients  of  all  the  co-functions  have 
a  negative  sign.     The  significance  of  this  will  be  seen  later. 


214  Elementary  Calculus. 

ii.   y=     -  tan  (nx).       12.   y=  -    —  cos5  (3  re). 

13.  y  =  cosn  x  sinn  x. 

14.  y  =  cot  x  -\-  J  cot3  re. 

.    tan3  re 

15.  y  =  re  —  tan  re  H . 


sin  x  cos  re 

17.  y  =  tan  re  (sin  re). 

18.  y  = 


19.  y  =  \/a  cos2  re  +  b  sin2  re. 

20.  y  =  sin  ax  (sin  re)a. 

Of  what  functions  are  the  following  the  differential  co- 
efficients: 

dy  •  4 

21.  -*—  =  5  sin4  re  cos  re. 

22.  -f-  =  a  [cos  (b  +  ax)  +  sin  (b  —  ax}], 
dx 

dy___ 
dx 

24.  -2-  =  —  20  x  cos4  2  re2  sin  2  re2. 

25.  2  m  cot  m/  cosec  wre. 


Elementary  Calculus.  215 

DIFFERENTIATION    OF    LOGARITHMIC    AND 
EXPONENTIAL    FUNCTIONS. 


The  series  y  =  A  +  B#  +  C*2  +  D^  +  .  .  . 
ART.  23.     Consider  the  geometric  series, 

i  +  i  +  (i)2  +  (i)3  +  .  .  . 

the  value  of  which  when  the  number  of  terms  is  infinite  is 
2.     We  can  approach  this  value  to  any  required  degree 
of  accuracy  by  taking  a  sufficient  number  of  terms. 
The  general  notation  for  such  a  series  is  as  follows: 

y  =  A  +  Ex  +  Cx2  +  D^3  +  ... 

when  A,  B,  C,  etc.,  are  constants.  The  calculation  of 
numerical  quantities  and  of  experimental  results  is  often 
referred  to  a  series  of  this  form. 

In  order  to  calculate  the  logarithms  of  numbers,  we 
make  use  of  a  series  in  which  x  either  is  equal  to  or  in- 
volves the  quantity  whose  logarithm  is  sought,  and  hence 
the  latter  can  be  calculated  to  any  required  degree  of 
accuracy. 

Such  a  series  to  be  of  practical  value  should  possess  the 
following  properties  :  it  must  converge  rapidly,  so  that 
it  will  not  require  a  large  number  of  terms  to  be  taken 
before  the  necessary  accuracy  is  reached,  and  it  must  be 
convenient  of  computation. 

The  binomial  theorem  supplies  us  with  an  expression 
of  the  form  y  =  A  +  Ex  +  Cx2  +  Dx?  .  .  .  ;  and  it 
has  been  found  that  the  determination  of  the  value  of 

(i  +  —  1  ,  when  n  becomes  infinite,  forms  a  suitable  start- 
n  I 

ing-point  from  which  to  begin  investigations  with  a  view  of 
obtaining  a  practical  logarithmic  series.  This  will  be 
discussed  in  its  proper  place. 


216  Elementary  Calculus. 

ART.  24.     The  'value  of    li-i  —  )    when  n  becomes  m- 

\        »/ 
•finite. 

I        i    \n 
Suppose  in  the  expression    [  i  +  --   1  we  put  n  =  oo.,  we 

\         n  I 

get      /i  +  -  -\  °°  =  (i  +  0)°°=    i00;  now  i°°  is  indeter- 

minate, for  infinity  has  no  definite  value;  we  regard  the 
symbol  oo  as  referring  to  a  magnitude  which  is  greater  than 
any  we  can  conceive. 

We  shall  refer  to  the  matter  of  indeterminate  forms  in  a 
subsequent  article.  In  the  mean  time  we  shall  show  that 
by  approaching  the  calculation  in  another  manner  we  can 

obtain  a  more  definite  result  for  the  evaluation  of  [  i  +  —  ) 

V         »/ 

when  n  =  oo  . 

By  the  Binomial  Theorem,  we  have 


1.2 

n  (n-i)  (n  -  2)  /A8 
1.2.3 


fa  ~  A       (n~  A  /^ 
i  +  i  +  V^-^  +V    n     A 


+    . 

1.2  1.2.3 

If  w  =  GO  ,  then  terms  such  as  -»    —  ,    etc.,  vanish; 


n     n 


1.2  1.2. 

=    2.71828       .... 

We  will  put  e  =  2.71828. 


Elementary  Calculus. 


217 


The  evaluation  of  e  to  any  required  degree  of  accuracy 
can  be  conveniently  performed  as  follows  : 


i  .000000 


2 

3 
4 

5 
6 

7 

8 

9 

i  .000000 

0.500000 

0.166667 

0.041667 

0.008333 

0.001389 

0.000198 

0.000025 

0.000003 

adding;        2.718281  =  e. 

Now  if  ax  =  N  then  loga  N  =  x.  If  then  we  can  obtain  a 
convenient  series  for  ex  we  shall  be  able  to  calculate  the 
logarithms  of  numbers  to  the  base  e;  for  if  ex  =  Nt,  then 
loge  Nt  =  x.  Let  us,  therefore,  endeavor  to  develop  a  series 
for  ex. 

ART.  25.     The  expansion  of  ex  and  the  logarithmic  series. 


i  H 


But 


T  \nx 

= 
n] 


=  ex. 


n  i 

nx  (nx—i]   (nx  —  2) 
1.2.3 

=  i  +x  +—  x 


!)  A 
\n  I 


218  Elementary  Calculus. 


+ 


+ 


I   +  X   + 


.        v         nxj  \         nx,    . 

Z3 

Now  if  n  =  oo  then  the  terms  — >     — >    etc.,  vanish. 
Hence  we  have 

xv.2  ^  ^4 

-«  I  I      *-v  i      »A/  i      ^  i 

Z2 

Now  put  jc  =  2  then 


Z3          /^ 

I 

1                I^ 

1.2.3 

1.2.3.4 

4- 

2-3-4-S 

=  i  +  2  +  2  +  1.333  +  o-667  .+  °-267   +  -  -  - 
=  7.266. 

Hence  we  have  log,,  7.266  =  2  nearly. 
It  is  obvious  that  the  above  series  would  be  far  from 
practical,  since  it  converges  slowly  and  it  would  be  diffi- 
cult to  obtain  the  logarithms  of  consecutive  integers.  It 
is,  however,  easily  possible  to  obtain  either  by  elementary 
mathematics,  or  by  an  application  of  the  calculus  (see 
Art.  54)  the  following  series, 

oc2         of  x4 

loge  (l    +  X)  =  X  -  -  H —    +       .... 

234 


Elementary  Calculus.  219 

This  is  known  as  the  Logarithmic  Series,  and  by  its  means 
we  could  calculate  many  logarithms,  but  since  it  also  con- 
verges slowly  and  only  between  the  values  x  =  +  i  and 
x=  —  i,  it  is  not  suitable  for  general  logarithmic  compu- 
tation. From  this  latter  series  we  can,  however,  obtain  the 
following: 

LO&  (Z  +  I) 

=   10&  Z    +  2    f  -  -  --  1  ---  -  --  1 

|_2Z  +  I  3    (2Z   +   I)3 


5(2Z+I)5 


7(2Z+i)7 

This  series  is  most  convenient  for  our  purpose,  for  in- 
stance if  Z  =  i,  then 

\oge  2  =  loge  i  +  2      -  H 1—  H ^—  +  .  . 

l_3       3  (3)3       5  (5)5  J 

.'.       Loge  2  =    0.6931. 

And  in  a  similar  manner  the  logarithms  of  other  quantities 
could  be  calculated. 

ART.  26.  The  logarithmic  modulus.  Logarithms  cal- 
culated to  the  base  e  are  known  as  Napierian  logarithms, 
because  of  their  introduction  by  Napier;  they  are  also  called 
Natural  Logarithms.  This  latter  term  was  applied  because 
they  appeared  first  in  the  investigation  conducted  for  the 
purpose  of  discovering  a  method  for  calculating  logarithms. 
The  base  e  is  used  exclusively  in  higher  mathematics,  but 
this  system  is  not  suitable  for  practical  computation;  the 
student  will  be  aware  that  for  the  latter  purpose  the  base 
10  is  chosen. 

We  will  now  show  how  logarithms  to  the  base  e  can  be 
transformed  to  the  base  10  and  vice  versa- 
Let  y  =  log,  x  and  z  =  Iog10  xt 
then                   ey  =  x  and  io2  =  x. 
.'.  e»  =  ioz. 


220  Elementary  Calculus. 

I.  To  transform  Iog10  x  to  log,  x,  we  had 

ey  =  io2. 

/.  y  log,  e  =  z  log,  io. 

But  log,  e  =  i  and  log,  io  =  2.30258,  and  since  y  =  log,  # 
and  z  =  Iog10  #, 

/.  \ogex=  2.30258  Iog10  #. 

The  quantity  2.30258  is  called  the  Modulus  of  the  Nap- 
ierian logarithms  and  is  often  denoted  by  M.  In  this 
notation  we  have 

log,  x  =  M  Iog10  x. 

II.  To  transform  log,  x  to  Iog10  x,  we  had 

ev  =  ioz. 

.'.  y  Iog10  e  =  z  Iog10  io. 

Now  y  =  log,  x  and  Iog10  e  =  0.43429,  while  Iog10  io  =  i 
and  z  =  Iog10  x. 

Hence  Iog10  x  =  0.43429  log,  x. 

The  quantity  0.43429  is  called  the  Modulus  of  the  Briggs 
System  and  is  denoted  by  m.  We  therefore  have, 

Iog10  x  =  m  log,  x. 

ART.  27.     The  relation  between  M  and  m. 
We  have     Iog10  x  =  m  log,  x  and  log,  x  =  M  Iog10  x. 

Now  loge  x  =  — B® —  • 

m 

Substituting  in  the  second  equation  above  we  get 

•2&*2  =  M  .  Iog10  x. 
m 

.*.  M  =  —  and  m  =  —— 

or  M  .  m  =  i. 

Hence  to  transform  logarithms  from  the  base  a  to  the 

base  b  multiply  by  ^ Note  log,  a  = 

loga  b  loga  e 


Elementary  Calculus.  221 

ART.  28.     The  d.  c.  of  y  =  loge  x.    We  will  now  write  Inx 
for  log,  x. 

We  have  y  =  Inx. 

.-.  y  +  Ay  =  ln(x 


Ay  =  ln(x  +  Ax)  —  Inx  =  In 
Multiplying  by  -  •    we  get, 


if)** 


Hence        -        =        te      1  +  ~ 


A     /y»  /y, 

If  L±X  becomes  dx  then  —  =  o,  while  -  -  =  oo  . 

x  &x 

Putting  -?-=  n  then  —  =  -  ,  and 

hx  x        n 

I     .    AJC\-^       /      .    i  \n 

[  I    H  --    A*  =       I    +   -  , 

V       *i      \      *) 

which  for  n  =  oo  is  equal  to  e  (Art.  24). 

Hence  we  get  -*-  =  —  /we, 

<fo       # 

but  Ine  =  i, 

<ty  __  i_ 
dx       x 

ART.  29.      The  d.  c.  oj  y  =  logax, 
y  =  loga*, 
/.  av  =  x. 
ylna  =  Inx, 


222  Elementary  Calculus. 


But  by  Art.  27,      — - —  =   loga  e. 
Jog,  a 

.-.     y   =  Inx  .  \ogae, 

and  -*-  =  —  loga  e. 

ax       x 

Note  loga  e  is  a  constant,    .*.     — 2£«_£  =  o,  hence  the  sec- 

dx 

ond  term  in  the  differentiation  of  the  product  is  zero. 

ART.  30.  Thed.c.o}y=a*  _ 

y  =  ax 
.-.  Iny  =  x  Ina, 

...  i.  .  •&   .  Ina. 
y       dx 

.    &-  =   anna, 
dx 

ART.  31.  The  d'  c"  °t  y  =  e* 

y  =  ex 

2 


1.2  1.2.3 


1.2.3.4 
Differentiating  each  term  we  get, 


1.2        1.2.3 

*+* 


0»  1.2  1.2.3 

Hence  ^-  =  ^. 


This  is  a  function  of  great  importance,  and  is  the  only 
one  known  whose  differential  coefficient  is  equal  to  the 
function  itself.  The  appearance  of  ex  and  eax  in  many 


Elementary  Calculus.  223 

physical  formulae  makes  these  quantities  of  particular 
interest  to  the  student,  who  will  have  no  difficulty  in  show- 
ing that  when  y  =  eax  then  -2-  =  aeax  by  a  process  similar 

dx 

to  the  above. 

ART.  32.     The   d.  c.  of  y  =  uv.     Let  y  =  uv  when  both 
u  and  v  are  functions  of  x. 

Iny  =  v  Inu. 

i      dy  i  du     .   7       dv 

.'.    -  .  -f-  =  v  .  - h  Inu  —  • 

y     dx  u  dx  dx 

If  we  now  multiply  by  uv  we  get, 

dy            „  n  du          „  7      dv 
-f-  =  m»-          +  uvlnu 

dx  dx  dx 

Hence,  to  differentiate  a   junction  of  the  form  y  =  uv\ 
first,  differentiate  as  though  u  were  variable  and  v  constant, 

(as  when  y  =  xn,  -*L  =  nxn~l) ;  second,  as  though  v  were 
(tx 

variable  and  u  constant  (as  when  y  =  ax,   -2-    =  axlnx) 

dx 

and  take  the  sum  of  the  results. 

The  following  table  gives  the  differential  coefficients  thus 
found : 


y=  logex 
y  =  log«; 

^      fc 

dx 

X 

I 

# 

y=  ax  ; 

dx 

=  a*  log,  a. 

y=ex; 

dx 

=  e*. 

y=f, 

dx 

=   ae0*. 

224  Elementary  Calculus. 

EXERCISE    IV. 

i.  y=  In  (2x2  -  i).  2.   y=  3 

3.  y  =  ex  .  xa.  4.   y  =  xx.         _ 

5.  y=ex  sin  x.  6.   y  =  aln  (\/x  +  a) 

7.  y=  alnx.  8.    y  =  cos  (Inx). 

9.  y  =  In  (Inx).  10.   y=  (ex)x. 


II.     ?=      **)*.  12. 

i  +  ex 
13.   y  =  eax  sin  nx. 


=  In  I1  +  *  ) 
\i  -  */ 


15- 


17.   y  =  log  cot  ex. 

19.   y  =  Inx  —  In  (a  —  \/a2  —  x2). 
ao.   y=ln   T  ~  cos  *  . 

I    +  COS  X 


21.   y  =  /»         +  *  +  V*2    +bx 


22.   y  =  as*nx.  23.   ^  =  ex  . 

24.  y  =x  wvw  (u,  v  and  w  are  functions  of  x). 

25.  y  =  eat  (cos  w#)*. 

III.  Differentiation  of  the  Inverse  Trigonometrical 

Functions. 

ART.  33.  When  we  wish  to  express  in  symbols  that 
y  is  an  angle  whose  sine  is  x,  we  write  y  =  sin-1  x,  and 
similarly  if  we  write  y  =  cos-1  x,  y  =  tan-1  x,  we  mean  that 
y  is  an  angle  whose  cosine  or  tangent  is  x.  Now  sin-1  J 
=  30°,  from  which  we  at  once  obtain  the  inverse  expres- 
sion sin  30°  =  J;  clearly,  if  y  =  sin-1  x  then  x  —  sin  y. 


Elementary  Calculus.  225 

The  German  mathematicians  write  y  =  arc  sin  x  instead 
of  y  —  sin-1  x.  The  former  expression  may  be  read  y  is 
an  arc  whose  sine  is  x.  A  similar  interpretation  is  given 
to  y  =  arc  tan  x  and  y  =  arc  sec  x,  and  so  on. 

The  inverse  trigonometrical  functions  y  =  sin-1  x, 
y  =  cos-1  x,  etc.,  are  of  great  importance  in  the  Integral 
Calculus. 

ART.  34.     The  d.  c.  of  y  =  sin~l  x  and  y  =  cos~lx. 

If  y=  sin-1  x, 

then  x  =  sin  y, 

and  dx  =  cos  y  .  dy  =  \/i  —  sin2  y  dy. 


Hence 

dx       vi  —  x2 

The  sign  of  the  root  depends  upon  that  of  cos  y  in  the 
expression  dx  =  cos  y  dy.  For  angles  in  the  first  quadrant 
this  is  clearly  positive. 

By  a  similar  process  the  student  will  find  that  if 

y  =   COS"1  X, 


then  =-     ,——• 

dx  vi  —  x 

ART.  35.  The  d.  c.  of  y  =  /aw-1  #  awe?  a?/-1  #. 

If  y  =  tan-1  x, 

then  #  =  tan  y, 

and  (foe  =  sec2  y  dy  =  (i  +  tan2  y)  dy. 

.  dy               i 


Hence 


dx        i  +  tan2  y 
dv  i 


Similarly,  if  y  =  cot-1  x,  -j-  —  -         — 2 


226  Elementary  Calculus. 

ART.  36.     The  d.c.  of  y  =  sec~lxandy  =  cosec*1  x. 
If  y=  sec~X 

then  x  =  sec  y, 


Hence 


doc  =  sec  y  tan  y  dy  =  sec  y  \/sec2  y  —  i  dy, 

dy_= * 

dx       sec  y  \/sec2  y  —  i 
i 


then 


In  like  manner,  if  y  =  cosec-t  x, 
dy  i 


x\/x2  -i  ' 


ART.  37.     The  d.c.  of  y  =  vers~l  x  and  covers-1  x. 
If  y  =  vers"1  x, 

then  x  =  vers  y  =  i  —  cos  y, 

dx  =  sin  y  dy  =  \/i  —  cos2  y  dy, 
dx  =  \A  —  C1  —  vers  ;y)2  cty 
=  \/2  vers  y  —  vers2  y  dy 
=  \/2  x  —  x2  dy. 

"     S"  = 


Similarly,  if  y  =  covers  -1  x,  then  -*-  =  - 


dx  \/2  x  —  x2 

Note  that  the  differential  coefficients  of  all  the  co-inverse 
functions  have  a  negative  sign,  and  that  in  each  case  where 
a  root  occurs  any  ambiguity  of  sign  may  be  disposed  of  by 
referring  to  some  previous  function  of  y. 


Elementary  Calculus.  227 

The  following  table  gives  the  above  results   in  concise 
form: 


=  sin-i  x.        dy_  =       - 


y 


dx      \A  -  & 

=  cos-.*;    |^=_-^==. 


-i  dy  i      . 

y  =  tan     *'        ^  ~~  i  +  x2 ' 


dy  i 

vV, 

y  =  sec-1  *; 


-y  =  cot-1  *; 

d»  i  +  x2 


y=  cosec-1*;     r*-  =  —        /-^ 

^*       *  v^  —  i 


!  ^  T 

y  =  vers-1  *; 

^          \/  2  X  —  X' 


covers 


-i    .     dy_  :     _       _] 


(^     "  \/2X~X* 

EXERCISE  V. 

i.   y  =  sin-1  (2  *).  2.    y  =  tan-1  3  a2. 

.  .  a  — 

3.  y  *  cos-1   -  •  4-  y  = 

x 


5.   y  =  cos~1\/a^.  6.   y  =  sin-1  \A 

tan-'-     8-    ==  tan"' 


g.   y  =  arc  sin  2  a*3.        10.   y  =  arc  tan      / — 


228 


Elementary  Calculus. 


ii.   y  =  b  arc  cot  —^  .         12.  y  =  a  .  sin-1  (     x     ) 
V  #  V*  ~  */ 


13-   y  =  b   sin-1  ^ 


cos  # 


1 6.   y  =  sec-1 
18.      =  e'n*. 


20.      =  covers-1  — 


22.       =  cot- 


>     ;y=   CQt 


17.   y  =  x  .  etaQ-^. 
19.   y  =  ex  sin-1  2  #. 


21.   y  =  vers-1-. 
x 


23.   y  =  arc  cos  \cos 


24.   y  =  arc  cos 
25-   y  =  i  cot"1 


2  3T  —   I 


CHAPTER  III. 
INTEGRATION. 

ART.  38.  In  Chapter  I  we  found  that  if  y  =  f(x)  be 
the  equation  to  a  curye,  then  the  Differential  Coefficient 

dy 

-*-  expresses: 

dx 

(1)  The  rate  of  change  of  the  function  as  compared  with 
the  rate  of  change  of  the  independent  variable. 

(2)  The  gradient  of  the  curve  at  any  point. 

Now  suppose  the  differential  coefficient  of  a  certain 
function  y  =  f(x)  be  given;  would  it  be  possible  to  obtain 
a  law  which  would  enable  us  to  find  the  original  function 
from  which  the  given  differential  coefficient  has  been 

derived?     For  example,  if  —  ==3  ax2  or  dy  =  3  ax2  .  dx, 

dx 

of  what  function  is  3  ax2  the  differential  coefficient? 
Let  us  examine  the  following  table: 

If   y  =  ax,       y  =  ax2,  y  =  ax3,  y  =  ax4  .    .   . 

then 

dy  =  a  dx,  dy  —  2  ax  •  dx,  dy  =  3  ax2  dx,  dy  =  4  ax3  dx. 

If  y=-x2,     y=-x3,         y=-x* 
2  3  4 

dy  =  ax .  dx,  dy  =  ax2  dx,      dy  =  ax3  dx 

(I)  Notice,  that  in  each  case,  if  we  multiply  the  differ- 
ential coefficient  by  x,  or,  what  is  the  same,  raise  the  power 
of  x  in  the  differential  coefficient  by  unity,  we  obtain  the 

229 


230  Elementary  Calculus. 

index  of  x  in  the  original  function.     (In  differentiating  we 
diminished  the  power  of  x  by  unity.) 

(II)  Again,  if  we  divide  by  the    increased  power  we 
obtain  the  numerical  factor  of  the  original  function  in  each 
case. 

(III)  The  constant  factor  a  remains  unaltered. 

(IV)  The  differential  disappears. 

Take  the  general  case,  —  =  axn  or  dy=  ax11  dx.    Apply- 
(toe 

ing  the  above  rules  we  obtain  the  original  function, 

xn+l 

Note  if  we  differentiated  this  latter  expression,  we  would 

have  -2-  =  — - —  (n  +  i )  x  "+1-1, 

dx       n  +  i 

and  hence,          dy  =  axn  .  dx. 

The  process  of  finding  a  function  when  its  differential 
coefficient  is  given,  is  called  Integration,  and  we  would  say 
in  the  above  case  we  had  integrated  the  expression  axn  .  dx. 

We  have  now  the  following  rule: 

To  integrate  a  differential  of  the  form  axn  dx,  first  raise  the 
power  of  x  by  unity,  then  divide  by  the  raised  power;  omit 
the  differential  of  the  variable. 

Example:     Suppose  dy  =  3  x15  dx. 

Integrating,  we  find  y  =  3  —  =  —  x1Q. 

10         16 

ART.  39.  It  was  supposed  by  Leibnitz,  that  a  function 
was  made  up  of  an  infinite  number  of  infinitely  small  differ- 
ences (differentials),  and  that  their  sum  made  up  the  func- 


Elementary  Calculus.  231 

tion.  Hence,  to  show  that  the  sum  was  to  be  taken,  the 
letter  S  was  used.  We  might  thus  write  S  dy  =  S  (3  x2  dx\ 
and,  therefore,  y  =  x3. 

Later,  for  convenience,  instead  of  the  letter  S  the  symbol 

I  was  employed.    This  symbol,  it  will  be  noticed,  is  simply 

an  elongated  S.  It  is  called  the  Integral  sign,  and  the 
process  which  it  represents,  Integration.  The  word 
''Integrate"  means  "to  form  into  one  whole,  or  to  give 
the  sum  total  of." 

In  modern  mathematics  we  would  write: 

Given  dy  =  3  x2  dx. 


read,   (The  integral  of  dy)  =  (the  integral  of  3  x2  dx). 


y  =  x3. 


Notice  that  the  integral  sign,    I  ,  is  only  a  symbol,  which 

can  be  looked  upon  as  meaning  that  we  are  to  find  the 
function  whose  derivative  with    respect  to  x  is  a  certain 


the 


given   quantity.     Thus    13   x2  dx  =  x?,  can  be  read, 

function  whose  derivative  with  respect  to  x  is  3  x2  dx,  is  x3. 

We  see  from  the  above  discussion  that  Integration  may 
be  looked  upon  as  the  inverse  of  Differentiation.  In  fact, 
problems  of  Integral  Calculus  are  dependent  upon  an 
inverse  operation  to  those  of  Differential  Calculus. 

ART.  40.  The  constant  of  integration.  Let  us  now 
take  the  equation  y  =  x2.  If  we  plot  the  corresponding 
graph  we  shall  obtain  a  curve,  known  as  a  parabola,  which 


232  Elementary  Calculus. 

will  cut  the  ^-axis  at  y  =  o;  from  the  equations,  y  =  #2  +  i, 
y  =  x2  +  2,  y  =  x2  +  3,  etc.,  and  again  y  =  x2  -  i, 
y  =  x2  —  2,  y  =  x2  —  3,  etc.,  we  obtain  a  series  of  similar 
curves,  with  coincident  axes,  which  will  cut  the  ^y-axis 
at  points  y  =  i,  y  =  2,  y  =  3,  etc.,  and  also  at  y  =  -  i, 
y  =  -  2,  y  =  -  3,  etc. 

A  general  expression  for  all  such  curves  would  be 
y  =  x2  +  C,  where  C  is  a  constant.  When  the  value  of 
C  is  known,  then  a  particular  curve  is  indicated. 

Let  us  take  the  differential  coefficient   -%-   =    2  x,  or 

dx 
dy  =  2  x  dx.     By  integration  we  have  from 

dy  =  2  x  dx, 
y=X\ 
But    —  =  2  x  would  be  obtained  by  differentiating  an 

(IX 

infinite  number  of  expressions  of  the  form  y  =  x2  +  C. 
There  is  nothing  to  tell  us  definitely  from  which  special 
function  the  2  x  has  been  obtained,  hence  we  see  that  we 
must  write: 

Given  -2-  =  2  x, 

dx 

or  dy  =  2  x  dx, 

then  I  dy  =       I  2  x  dx, 

and  y  =  x2  +  C. 

C  is  called  a  constant  of  Integration,  and  must  always  be 
added  when  integrating  an  expression  about  which  nothing 
more  is  known  than  that  it  is  the  differential  coefficient  of 

a  certain  junction.    An  expression  such  as  /  2  x  dx  =  x2  +  C 


Elementary  Calculus.  233 

is  called  an  Indefinite  Integral,  because,  from  the  given 
data,  the  function  cannot  be  definitely  determined.  In 
practical  problems  we  can  generally  obtain  one  or  more 
conditions  which  will  indicate  the  required  functions. 

Suppose,  for  instance,  we  had  given  dy  =  2  x  dx  and  the 
condition  that  the  curve  pass  through  the  point  x  =  2, 

y=  5- 

We  have  by  integration,    y  =  x2  +  C. 
.'.  substituting,        5=4  +  C, 
and  C  -  I. 

Hence  the  function  is  definitely  found  to  be  y  =  x2  -f-  i. 
This  expression  obtained  from  the  Indefinite  Integral  is 
called  a  Definite  Integral. 

Take  dv  =  a  dt. 

Here  idv=  Cadi. 


/*-/. 


/.  v  =  at  +  C 

where  a  is  the  original  acceleration,  due  to  gravity,  and 
C  the  constant  of  integration.  Now  if  the  condition  is 
imposed  that  the  body  starts  from  rest,  when  t  =  o, 
v  =  o,  and  .'.  C  =  o,  and  we  get  the  definite  integral 
v  =  at,  where  C  stands  for  the  initial  velocity,  which  is 
zero  in  this  case. 

From  the  above  we  see  that  strictly, 

axn  dx  =  a  4-  C, 


and  therefore,         /  3  x4  dx  =  $-  x5  +  C. 
J  5 

In  practice,  however,  the  constant  of  integration  is  often 
understood.  We  shall  refer  again  to  the  integration  con- 
stant in  a  later  article. 


234  Elementary  Calculus. 

ART.  41.  A  constant  factor  may  be  placed  outside  the 
integration  sign.  The  differential  of  ax  is  a  dx, 

hence,  I  a  .  dx  =  ax  —  a    I  dx. 

Rule.  If  an  expression  to  be  integrated  has  a  constant 
factor,  this  factor  may  be  placed  without  the  integration 
sign. 

ART.  42.  The  integration  of  a  sum  or  difference.  In  the 
Differential  Calculus,  we  found 

d  (u  j:  v  jb  w)  __  du        dv     .   dw 
dx  dx        dx       dx  ' 

or  d  (u  ±  v  ±  w)  =  du  ±  dv  ±  dw, 

hence        /  (du  ±  dv  ±  dw)  =    I  du  ±      I  dv  ±      I  dw. 

Rule.  The  integral  of  an  algebraic  sum  is  equal  to  the 
algebraic  sum  of  the  integrals  of  the  various  terms. 

ART.  42 a.  A  problem  of  integral  calculus  geometrically 
considered.  Mechanics  supplies  us  with  the  following 
relation : 

v  =  at 

where  v  =  velocity,  a  =  acceleration,  and  /  =  time.  In 
Chapter  I  we  realized  that  v  =  -^-  where  s  =  space  trav- 
ersed in  the  time  t. 

Hence  ~  =  at, 

at 

and  ds  =  at  dt. 


.'.  ids  =   i  at  dt. 


.'.  s  =     at\ 


Elementary  Calculus. 


235 


ds 


We  have  thus  found  that  the  differential  coefficient  — =  at 

dt 

results  from  the  differentiation  of  the  function   5  =  \  at2. 

We  will  now  investigate  this  matter  geometrically  and 
the  student  will  at  once  be  convinced  that  the  Integral 
Calculus  has  a  much  wider  scope  than  has  been  thus  far 
indicated. 

The  graph  of  v  =  at  is  a  straight  line,  and  since  we  will 
assume  that  there  is  no  initial  velocity,  and,  therefore,  no 
added  constant,  this  straight  line  passes  through  the  origin. 


In  Fig.  1 6  let  OA  represent  the  graph  of  v  =  at,  while 
the  units  of  time  and  velocity  are  referred  to  the  co-ordinates 
as  shown. 

Suppose  the  time  represented  by  OB,  which  is  the 
abscissa  of  any  point  A,  to  be  divided  into  a  number  of 
equal  parts,  and  the  construction  of  the  figure  completed 
as  shown.  In  the  case  of  uniform  velocity  s  =  vt. 

Take  any  small  time  interval  CD  and  suppose  the 
velocity  of  the  moving  body  constant  jor  this  short  period. 
The  velocity  of  the  body  at  the  beginning  of  this  time 
interval  would  be  represented  by  CE  and  at  the  end 
bvDH. 


236  Elementary  Calculus. 

Since  s  =  vt  is  the  space  traversed  by  the  body  during 
the  time  represented  by  CD,  then,  under  the  supposition, 
that  throughout  this  short  time  interval  a  constant  velocity 
equal  to  CE  is  maintained,  CE  X  CD  or  the  area  of  the 
rectangle  CDFE  would  geometrically  represent  the  space 
traversed. 

Again,  since  DH  represents  the  final  velocity  at  the  end 
of  the  time  interval  CD,  then  the  area  of  the  rectangle 
CDHG  would  represent  the  space  traversed,  under  the 
supposition  that  throughout  the  time  CD  this  latter  velocity 
be  constantly  maintained.  The  actual  space  traversed 
would  be  more  than  the  first  result  would  indicate,  and 
less  than  the  latter. 

Now  the  complete  space  traversed  would  be  clearly  mere 
than  that  represented  by  the  shaded  rectangles  and  less 
than  that  indicated  by  the  larger  rectangles,  of  which 
CDHG  is  a  representative.  The  difference  or  error  would 
be  given  by  the  sum  of  the  small  rectangles,  one  of  which 
is  EFHG. 

Now  the  sum  of  these  latter  is  equal  to  the  rectangle 
D'BAK'.  But  the  area  of  D'BAK'  can  be  infinitely  reduced 
by  making  the  time  interval  small,  and  when  the  latter  is 
dt  or  infinitely  small,  the  area  of  D'BAK/  is  evanescent.  In 
this  case  the  error  or  difference  disappears  and  the  whole 
space  traversed  during  the  time  OB  is  represented  by  the 
area  of  the  triangle  OAB. 

Now  the  area  of  the  triangle  OAB  =  J  .  OB  X  BA. 

But  OB  =  /  and  BA  =  v. 

.  Hence  OAB  =  \t.v=  J  / .  a/, 

or  area  of  OAB  =  J  at2. 

But  the  area  of  OAB  represents  s, 
.-.$=*  at\ 


Elementary  Calculus. 


237 


Hence  we  find   that  when  we  integrate  thus,   I  d s  = 
at .  dt,  and  find  5  =  J  a/2,  we  have  really  obtained  the 


sum  of  an  infinite  number  of  elementary  areas,  each  v  .  at 
or  at  .  dt,  the  total  of  which  gives  the  space  traversed  by 
the  body  during  the  time  /,  and  moving  in  accordance  with 
the  law  v  =  at. 

The  summation  of  elementary  areas  with  a  view  of 
obtaining  a  result  indicated  by  their  total  is  a  marked 
feature  of  the  Integral  Calculus. 

ART.  43.  The  definite  integral.  Should  it  be  required 
to  determine  the  space  traversed  by  a  moving  body  under 


the  law  v  =  at  during  a  finite  time  interval  CD  we  might 
proceed  thus:  putting  OD  =  /2  and  OC  =  ^  (Fig.  17),  and 

integrating     I  ds  =   I  at  .  dt,  we  get  s  =  J  at2  +  C,  as  we 

have  already  seen,  and  if  the  initial  velocity  is  zero  we 
have  5  =  i  at2. 
The  space  traversed  from  zero  to  /2  is  represented  by  the 


238  Elementary  Calculus. 

area  of  the  triangle  ODH  =  \  at22,  and  that  from  zero  to 
*!,  by  the  area  of  OCE  =  \  at2. 

Subtracting,  we  have  \  at2  —  ^  at2  =  area  CDHE,  which 
gives  the  required  space  traversed.  In  the  language  of  the 
Integral  Calculus  we  express  the  above  as  follows  : 

I    2  atdt  =      I  at2dt  —   I  a^  dt  =  J  at22  —  \  at2, 
or  thus, 


The  integral  I  2atdt  is  called  a  Definite  Integral;  /2-and 
Jt, 

/!  are  referred  to  as  the  superior  or  upper,  and  inferior  or 
lower  limit,  respectively.  We  read  the  expression  thus:  the 
integral  from  /t  to  /2  of  at  .  dt. 

It  will  be  noticed  that  the  quantity  enclosed  in  brackets 
is  the  solution  of  the  general  or  indefinite  integral,  and 
that  the  solution  of  the  definite  integral  is  obtained  by  sub- 
stituting first  the  upper  limit,  then  the  lower,  and  taking 
the  difference. 

The  constant  is  clearly  made  to  disappear  by  taking  the 
difference  between  the  integrals  formed  by  giving  two 
successive  values  to  the  independent  variable. 

To  find  the  value  of  a  definite  integral  solve  the  general 
integral,  then  substitute  first  the  upper,  then  the  lower  limit, 
and  take  the  difference.  This  process  will  be  made  clear 
by  the  following  simple  example: 

Required  the  space  traversed  between  5th  and  ;th  seconds, 
given  the  acceleration  equal  to  4  feet  per  second  per  second. 

5=    C7at.dt=[^  at2].7. 

Js 

.'•  *=  i-4-  (7)2-i-4.(5)2=48sq.ft. 


Elementary  Calculus.  239 

INTEGRATION  OF  GENERAL  FORMS. 

ART.  44.    It  is  to  be  observed  that  in  the  formula, 


aocn  dx  =  a  -       .....     (A) 

n  +  i 

x  stands  for  any  expression  whatever.  Hence,  whenever 
we  have  a  quantity,  monomial  or  polynomial,  raised  to  any 
power  and  the  differential  of  this  quantity  (without  its 
exponent),  formula  (A)  applies. 


Example.        I  (2  x3  —  3  x2  +  5)*   (x2  -  x)  dx  = 


what? 


Since  a  constant  does  not  affect  differentiation,  it  does  not 
affect  integration,  so  that  we  are  always  at  liberty  to  intro- 
duce a  constant  factor  behind  the  integral,  if  at  the  same 
time  we  divide  the  integral  by  the  same  factor,  in  order 
that  the  value  be  not  altered.  But  no  expression  contain- 
ing the  variable  can  be  removed  from  behind  the  integral  or 
introduced  in  any  way. 

In  the  example  above, 

d(2  x3  -  3  x2  +  5)  =  (6  x2  -  6  x)  dx  =  6  (x2  -  x)  dx. 
Hence  if  the  expression  (x2  —  x)  dx  be  multiplied  by  6,  it 
becomes   the   differential   of   2  x3  —  3  x2  +  5   and  we  get 
form  (A);  thus, 


f 


(2  Xs  —  3  x2  +  5)e  (x2  -  x)  dx  = 

r*-3*2  +5)i(6*a-6*)<fc;  = 
[Like  (A)],  [where  2=2^-3^  +  5]. 

2  ^  -  3  x2  +5)3  (*2  -  x)  dx 

^  (2X3-  ix2  +  $Y*   =   (2^-3^  +  5)', 
f  15 


240  Elementary  Calculus. 

f*      xdx 

Agaln 


xdx 
\/r2  -  x2  = 

-  x2)-*  (-  2  xdx)  =  -  (r2  - 


since      —  2  #  <fo  =  d(r2  —  x2). 

TRIGONOMETRIC   INTEGRALS    AND    LOG 
INTEGRALS. 

ART.  45.  Since  integration  is  the  reverse  of  differen- 
tiation, we  easily  derive  the  following,  by  reversing  the 
formulae  for  differentiation: 

/  cos  x  dx  =  sin  x  +  c. 
I  sin  x  dx  =  —  cos  x  +  c. 
I  sec2  x  dx  =  tan  x  +  c. 
I  esc2  x  dx  =  —  cot  x  -f  c. 
I  sec  x  tan  x  dx  —  sec  x  +  c. 

I  esc  x  cot  #  cfo  =  —  csc  x  -{-  c. 

dx 

s111"1  *  +  CJ  or  ~  cos-1  x  4-  c. 


/~   — 
V  I  —  # 


Elementary  Calculus.  241 

f    dx      =  tan-1  x  +  c. 
J  i  +*2 


\/a2  —  # 


-  sin-1*  +  c  or  -  cor**  +  c. 


— ^—  =  -  tan-1  -  +  cor  -  -  cot-1  -  +  c. 
+  x 2        a  a  a  a 

I  -  -  =  log  x  +  c,  etc. 
t/    ^c 

Put  these  all  into  rules. 

EXERCISE  VI. 

Integrate: 

i.       I  x%dx.  2.     I  (x  —  2)2  dx. 

4.  f(2  ^2  -  4  *  +  5)*  (x  -  i)  <&. 

5.  I  (jc2  —  i  )2  A;  dx.  6.     /  (jc2  +  3  #)2  ^- 
7-       /  (5^  -  3^cJ  +  i)^.   8.     I-        1-dx. 

/f  * — : •  10.    /  — ~  2*          dbe. 

(^2  +  i)J  J  ^2 


u 


.     J  (i  -  *)3  \/^  ^-       12.  J  (\/n  -\/x)2  dx. 
13-       f(3  *2  -  «*)*  (2  *  -  *2)  ^- 


2  Elementary  Calculus. 

C  dx  r          3/- 

14.       I   i  / I1?-    I  (i  —  V  jc)3  dx. 

J  3/x2  J 

16.  I  cos3  #  sin  #  dx.  17.  I  (i  —  cos  x 

18.  i  tan*  x  sec2  #  <fo.  19.  /  cot3  x  esc2  #  dx. 

20.  /  sec2  rv  tan  xdx.  21.  I  esc3  jc  cot  jc  c?^. 

22.  I  sin*  ^  cos  ac  dx.  23.  I  e^2  A;  dx. 

24.        I  tan  5f  dx.  25.    /  - 

J  J  sin  x 

I  cos  #2  rv  dx.  27 

/JC<fo 
^2  +  I  • 

J^*?dx 
JC  +  I 

2^  +  3 

34.  r-^fc.. 

t/    I    +  COS  X 


cos  x 


26. 

28 

30- 
32 


.    1 


p  sec2  x 
J        tan  3 


ry  dx 
tan  # 


ART.  46.  77ze  w«e  curve;  harmonic  motion.  Suppose 
P!  (Fig.  18)  is  a  body  moving  in  a  circle  with  uniform 
velocity,  the  centre  of  the  circle  being  O ;  let  P2  be  a  second 
body  moving  in  the  fixed  diameter  AB,  but  in  such  a  man- 
ner that  P2  always  maintains  a  position  at  the  foot  of  the 


Elementary  Calculus. 


243 


perpendicular  from  Pj  upon  AB.     Now  the  body  P2  travels 
backwards  and  forwards  upon  the  diameter  and  its  velocity 


A  X 


Fig.  18. 


will  be  at  a  maximum  as  it  passes  O  and  diminishes  as  it 
approaches  B  and  A;  such  motion  executed  by  P2  is  called 
Simple  Harmonic  Motion. 

The  distance  from  O  to  A  or  B  is  called  the  Amplitude. 
If  we  fix  upon  any  point  in  AB,  then,  once  at  each  complete 
revolution  of  P15  the  body  P2  will  pass  this  fixed  point, 
travelling  in  the  same  direction.  The  time  thus  occupied  by 
P2  in  completing  such  a  cycle  of  motion  is  called  a  Period. 
The  motion  of  a  tuning  fork,  an  oscillating  pendulum  and 
an  alternating  current,  are  good  examples  of  periodic 
motion.  The  change  of  position  or  motion  of  the  particle 
P2  is  clearly  a  function  of  the  time,  and  further  since  each 
cycle  of  motion  recurs  periodically,  we  say  that  the  Simple 


244  Elementary  Calculus. 

Harmonic  Motion  of  a  point  is  a  periodic  function  of 
time. 

In  general  a  Periodic  Function  is  one,  the  value  of 
which  recurs  at  fixed  intervals,  while  the  variable  increases 
uniformly. 

In  Fig.  1  8,  suppose  OP  is  a  revolving  radius,  and  tracing 
a  constantly  increasing  angle,  a. 

Putting  the  radius  of  the  circle  equal  to  unity 

then  sin  a.  =  P2Pi, 

or  in  general  y  =  sin  a. 

.'.  y  =  sin  (a  +  2  TT). 

Evidently,  then,  y  =  sin  a.  is  a  periodic  function,  and 
the  period  is  the  time  taken  to  complete  one  revolution. 
This  is  equal  to  271  divided  by  the  angular  velocity,  which 

we  will  call  6.  We  thus  have  the  Period  T=  —  • 

6 

The  Frequency,  or  the  number  of  periods  in  a  second,  is 


Note  that  0  =  2  n  .  —  ,  and  /.  6  =  2  nf. 

In  electrical  work  the  number  of  alternations  per  minute 
is  often  used  instead  of  the  frequency.  From  the  annexed 
diagram  it  will  be  seen  that  the  motion  of  the  Point  P3  is 
exactly  similar  to  that  of  P2,  excepting  that  when  P2  is  at 
the  extremity  of  its  path,  where  the  instantaneous  velocity 
is  zero,  the  point  P3  is  passing  through  the  O  with  its  maxi- 
mum velocity  and  so  on. 

Calling  the  radius  of  the  circle  a  (the  Amplitude),  we  have, 


Elementary  Calculus.  245 


but  cos  (90°  —  a)  =  sin  a, 

y  =  a  cos  (90  —  a). 

.  *.  y  =  a  sin  a, 
or  y  =  a  sin  (a  +  2  TT). 

Hence  we  see  that  y  =  a  sin  a  represents  the  Simple 
Harmonic  Motion  of  the  point  P3;  where  a  is  the  Ampli- 
tude and  a  the  angle  described  from  a  fixed  starting 
point,  and  is  the  product  of  the  angular  velocity  and  time, 
a.  =  dtj  we  generally  write  y  =  a  sin  dt. 

Note  that  since  the  sine  can  never  be  greater  than  +  i  or 
less  than  —  i,  hence  the  maximum  and  minimum  values 
of  sin  6t  are  +  i  and  —  i,  respectively. 

We  will  now  draw  a  graph  of  the  Simple  Harmonic 
Function  y  =  sin  a: 

If  a  =  o     y  =  o  a  =  «  —      y  =  0.707 


a  =  —     y  =  0.707       a  =  TT         y  =  o 
4 


a=-y=i  a=5JE       <y  =  —  .707 

2 


-         y  =  —  .707 

4 


o. 


246 


Elementary  Calculus. 


Referring  a,  expressed  in  radians  to  the  #-axis,  and 
using  the  same  scale  as  the  ordinate,  we  obtain  a  sinuous 
or  wavy  curve,  known  as  the  Curve  of  Sines  or  the  Har- 
monic Curve.  If  the  motion  of  the  point  giving  rise  to 
this  graph  be  made  quicker  or  slower,  the  undulations  of 
the  curve  will  be  more  widely  spread  or  brought  nearer 
together. 

Increase  in  Amplitude  gives  increased  rise  to  the  undu- 
lations and  vice  versa. 

Fig.    (i8a)   shows   the   same   curve   plotted  by   another 


Fig.  i8a. 


method;  the  student  should  have  no  difficulty  in  understand- 
ing the  principle  after  an  inspection  of  the  figure.  It  will 
be  noticed  that  the  curve  does  not  begin  upon  the  *-axis, 
but  that  the  periodic  time  is  counted  from  the  instant  that 
the  point  Pt  has  passed  through  the  angle  e.  This  angle 
e  is  called  by  electrical  engineers  the  lead;  when  negative 
it  is  known  as  the  lag. 

The  term  Phase  is  used  to  denote  the  interval  of  time 
that  has  elapsed  since  the  point  P  passed  through  its  initial 
position  at  A,  and  hence  e  is  often  called  the  Phase  Con- 
stant. 


Elementary  Calculus. 


247 


ART.  47.  Plane  areas.  Let  y  =  f(x)  be  a  curve,  and 
AB  a  fixed  ordinate.  Now  suppose  CD  =  y  be  a  second 
ordinate  corresponding  to  the  value  x  =  OC  (Fig.  19). 


Consider  the  area  ABDC,  call  this  area  u,  let  CF 
then  Aw  =  CFHD,  and  Ay  =  GH. 
Now  CDGF  <  Aw  <  CEHF  ;  but  CDGF  =  y  . 

and  CEHF  =  FH  .  A*. 

Hence  y  .  Ax  <  Aw  < 

AM 


, 
.'.  y  < 


.  T.TT 
<  FH. 


Now  the  smaller  A#  becomes,  the  more  nearly  will  y 

and  FH  approach    —  —  in  value;  hence  when  A#  becomes 
UkX 

dx,  then  FH  =  y  =   —   and  du  =  y  .  dor. 
(too 

Hence  if  any  area  is  bounded  by  a  curve  (y  =  /(#)),  a 
portion  of  the  abscissa,  and  two  ordinates,  then  the  differen- 


248  Elementary  Calculus. 

tial  of  such  area  (du)  is  equal  to  the  product  of  the  termi- 
nating ordinate  (y)  and  dx. 

Adopting  the  notation  of  the  last  paragraph  we  have, 
for  the  Definite  Integral  which  expresses  the  area  bounded 
by  the  curve,  part  of  the  abscissa,  and  two  ordinates,  a 
and  b,  this  expression 

Xb 
y  .dx. 
. 

Xb 
f(x)dx. 

NOTC:  y  .  dx  gives  a  numerical  measure  of  an  area 
which  may  be  found  as  follows: 

(I)  Integrate   the  given    differential  expression,   or  as 
we  say  find  the  indefinite  integral. 

(II)  Substitute  the    given  limits,  first  the  higher,  then 
the  lower;  subtract  the  latter  resulting  expression  from  the 
former. 


CHAPTER   IV. 

TANGENTS,    SUBTANGENTS,    NORMALS   AND 
SUBNORMALS. 

ART.  48.  In  Analytic  Geometry  it  was  found  that  the 
form 

y-y=m(x-oc'} (C) 

expressed  the  equation  of  a  straight  line  in  terms  of  its 
slope  (m)  and  a  fixed  point  (V,  y'). 

As  any  curve  may  be  regarded  as  generated  by  a  point 
moving  according  to  a  definite  law,  expressed  by  its  equa- 
tion, the  direction  of  a  curve  at  any  point  is  the  direction  in 
which  this  point  (taken  as  the  generating  point)  is  moving 
at  the  instant.  But  the  generating  point  if  not  constrained 
to  move  in  the  curve,  would  at  any  instant  move  off  in  a 
straight  line  (by  the  first  law  of  motion)  and  this  straight 
line  would  be  tangent  to  the  curve  at  the  point  of  departure; 
hence : 

The  slope  of  a  curve  at  any  point  is  the  slope  of  its  tan- 
gent at  that  point,  slope  meaning  as  usual  the  tangent  of 
the  angle  made  with  the  rv-axis. 

In  equation  (C),  if  (xf,  /)  is  a  point  on  a  given  curve, 
and  m  is  the  slope  of  the  tangent  at  that  point,  then  (C) 
is  the  equation  of  the  tangent  at  (xft  /).  But  if  y=  f  (x) 
(where  /  (x)  is  any  expression  containing  only  x  and 
known  quantities)  is  the  equation  to  a  curve  it  has  been 

shown  that  --21  —  the  slope  of  the  tangent  to  the  curve,  and  if 
dx 

the  coordinates  of  a  definite  point  on  the  curve,  like  (xf,  y'\ 

249 


250  Elementary  Calculus. 

be  substituted  in  the  value   of  — ,  it  will  then  represent 

dx 

the  slope  of  the  tangent  at  that  point;  say  ( -j- \=  slope  of  the 

\dx  I  xf,i/f 


tangent  at  (xf,  y'). 
Then  (C)  becomes 


x',  y' 

which  is  clearly  the  tangent  equation  at  (#',  /). 

ART.  49.  From  these  considerations  an  expression  for 
the  subtangent  is  readily  found,  in  exactly  the  same  way 
as  described  in  Analytic  Geometry  (see  Art.  50). 

Since  the  normal  is  a  perpendicular  to  the  tangent  at  the 
point  of  tangency  (#',  y'),  its  equation  will  be, 

i 

y  -  yf  =  -  (T-)     (*-*')•••  (N) 

\dx]#,j 

by  the  relation  between  the  slopes  of_J_  lines  as  developed 
in  Analytic  Geometry. 
This  equation  may  be  written: 


if  we  understand    -^—  to  represent  the  reciprocal  of  -2-  • 
dy  dx 

As  in  the  case  of  the  subtangent  the  subnormal  is 
readily  found  by  determining  its  ^-intercept  from  its  equa- 
tion (N). 


Let 

then 

whence 


Elementary  Calculus, 
y  =  o  in  (N), 


X  =   X*  + 


OC    .     (Fig.  20) 


B  0 


Fig.  20. 

But  subnormal,  BC  =  OC  -  OB  [P  =  (V,  /)] 


-^  -/(?•)    . 

J        \dxlx>,y> 


Corollary  :  The  lengths  of  tangent  ajid  normal  are 
readily  found,  since  they  are  the  hypotenuses,  respectively, 
of  the  triangles  APB  and  BPC. 


=  AB2+PB2=/2  (— 


V2 


^y  i 

dyJx',y>] 


and        PC2=PB2+  BC2  =  /2     i  + 


=  /2  fi  +  I^-Y        ] 
L         \<bj  «MfJ 


Example  :  Find    equation    of    tangent,  subtangent    and 
subnormal  to  the  ellipse  16  x2  +  25  y2  =  400  at  (3,  3J). 


252  Elementary  Calculus. 

From  1  6  x2  +  25  y*  =  400 

dy  =     _  16  x 
dx  2$y 

At  the  point  (3,  aJ)  this  becomes, 

m  .    16X3   =  -3- 

V**  /*>.  y'  25  X  V          5 

Hence  tangent  equation  is 

)  =  (3,3t)l 


or  5  y  +  3  ^  -  25  = 

^  \  i 

also  =  —  -  =  — 


/^  \  i 

= 
\dy  jx',y>       - 

Hence  subtangent  =  /(f  )^,  =  (f  )(  ~  | 


and  subnormal  =  /  (&\    .     -  l!  (  -  l)  -  -  A. 

\dx)x,,y'         5     \       S/  25 

ART.  50.  Subtangent,  subnormal,  etc.,  in  polar  co-ordi- 
nates. 

Using  the  Polar  System,  subtangent  and  subnormal  are 
denned  as  follows: 

The  subtangent  and  subnormal  are  respectively  the  dis- 
tances cut  off  by  tangent  and  normal  from  the  pole  on  a 
line  drawn  through  it  J_  to  the  radius  vector  of  the  tan- 
gency  point,  as  OT  and  ON  (Fig.  21). 


Elementary  Calculus.  253 

Calling  the  angle  TPO  between  radius  vector  and  tan- 
gent,  </>,  we  have  in  the  right  traingles  OPT  and  OPN, 


Fig.   21. 


subtangent,  OT  =  OP  tan  TPO  =  p  tan  (p.     Subnormal, 
ON  =  OP  tan  OPN  =  p  cot  (b  (since  OPN  =90°  -  TPO). 

The  angle  (p  is  determined  thus: 

Let  ACE  be  any  curve  (Fig.  22),  the  co-ordinates  of  C 


A/7 


Fig.  22. 


being  (p,  0),  and  of  A  being  (p  +  A/B,  6  +  A  6).    Then 


AB=  ApandAOC 


TanBAC=  —    [since    A0 
AB 


254  Elementary  Calculus. 

is  a  very  small  angle  the  arc   BC  does  not  differ  sensibly 
from  a  tangent  at  B,  say].     Whence 

tan  BAG  = 


(arc  BC  =  pA#,  since  an  arc  =  its  angle  multiplied  by  the 
radius).  As  the  point  A  approaches  C,  the  secant  AC 
approaches  the  position  of  a  tangent  at  C  (FG)  and  BAG 
approaches  the  value  (p  (OCG),  hence,  finally, 

,       pdd 
tan  0  =  c  -- 
dp 

TO 

Hence  polar  subtangent  =  p  tan  <p  =  p2  -  —  , 

dp 

and  polar  subnormal  =  p  cot  <b  =  -~  • 

do 


EXERCISE  VII. 

1.  Find  the  length  of  tangent  and  normal  for  the  para- 
bola y2  =  16  x  at  x  =  4. 

2.  Find  the  length  of  subtangent  and  subnormal  to  the 
ellipse  9  x2  +  16  y2  ==  144  at  (6,  6  Y/3)- 

3.  Find    the     equations    of    tangent    and    normal     to 
y2  =  16  x3  at  (i,  4). 

4.  Find  the  length  of  the  normal  to  x2(x  +  y)  =  4  (#  —  y) 
at  (o,  o). 

5.  Find  where  the  tangent  to  yax  =  x3  —  a3  is  parallel 
to  the  #-axis. 

6.  Find  where  the  normal  is  JL  to  the  rv-axis  on  the  curve, 
f  =  ^  (8  -  a;). 

7.  Find    the    angle    at    which    x2  =  y2  +  9    intersects 
4  ^2  +  9  f  =  36- 


Elementary  Calculus.  255 

8.  In    the    equilateral    hyperbola  x1  —  y2  =  16.     The 
area   of    the    triangle  formed  by    a   tangent  and  the  co- 
ordinate axes  is  constant  and  equal  to  16.     Prove  it. 

9.  At  what  angle  do  y2  =  8  x  and  x2  +  y2  =  20  intersect? 

10.  Show  that  the  subtangent  to  the  parabola  y2  =  2  px 
is  twice  the  abscissa  of  the  point  of  tangency. 

11.  Show  that  in  a  circle  the  length  of  the  normal  is 
constant. 

12.  The  equation  of  the  tractrix  being 


show  that  the  length  of  the  tangent  is  constant. 


CHAPTER  V. 
SUCCESSIVE    DIFFERENTIATIONS. 

ART.  51.     Since  -*—  is,  in  general,  purely  a  function  of 
dx 

x,  its  differential  coefficient  may  be  found  as  readily  as  that 

of  the  original  function.     It  is  usually  symbolized  thus,  —  ^  • 

dx 

For  example,  if  y  =  3  x3  +  2  x2  —  5  x*, 

JL=QXi  +  4X-s  x-l, 
dx 


2 

dx2 

Likewise  the  differential  of  this  second  differential  may  be 

found  in   the  same  way,   and  is  symbolized  as  -^-;  the 

dx? 

fourth  differential  coefficient  as  -^  ;   the  nth  as  —  -  .    it 

dx*  dxn 

sometimes  happens  that  the  successive  differential  coeffi- 
cient may  be  written  by  analogy  after  three  or  four  have 
been  found.     For  example  : 
y=  xm, 


dx 


256 


Elementary  Calculus.  257 

dn/v 

—  2—  =  m  (m  —  i)  (m  —  2)  .  .  . 

dxn 

(m  —  n  +  i)  xm~n 

If  the  function  be  an  implicit  function  of  x  and  y,  it  is 
not  necessary  to  put  it  in  explicit  form,  as  the  previously 
found  derivatives  may  be  used  to  find  successively  each 
higher  one.  For  example: 

x*  +  y2=r*      .     .     .     .     (i) 

dy 

Take  ^-derivative  :    2  x  +  2  y  -f~  =  o      ....     (2  ) 

ax 

solving  for    -%-  ,  -2-  =  ——...     (3) 

ax  ax  y 


substituting  value  of  -%-  already  found  from  (3)  in  (4), 
ax 


dx2 

_   3,2      2. 


da?  f  y5 

MACLAURIN'S   AND    TAYLOR'S    FORMULAE. 

ART.  52.  It  is  frequently  useful  for  purposes  of  calcula- 
tion to  express  the  value  of  a  function  in  the  form  of  a 
series.  For  example,  in  algebra,  the  binomial  theorem 
enables  us  to  develop  a  binomial  raised  to  any  power  into  a 
series  of  powers  of  the  single  quantities  involved,  as, 

(a  +  b)4  =  a4  +  4  a3  b  +  6  a2  b2  +  4  a  b3  +  b\  etc. 


258  Elementary  Calculus. 

Likewise  the  logarithms  of  numbers  and  the  trigonometric 
functions  are  computed  from  series. 

Hence  a  general  method  for  the  expression  of  any  func- 
tion of  x,  say,  in  series,  would  prove  exceedingly  useful. 

But  such  a  series  has  utility  only  when  its  sum  is  a  finite 
quantity.  In  general,  series  have  an  unlimited  number  of 
terms,  and  clearly,  unless  the  sum  of  these  terms  is  a  finite 
quantity,  it  is  utterly  useless.  A  series  whose  sum  is  finite 
is  called  a  convergent  series. 

It  is  only  with  such  series  that  we  shall  deal  here.  Let 
it  be  required  to  develop  f(x)  into  a  series  of  powers  of 
(x  —  m)  say.  Supposing  such  a  development  possible,  let 

/(*)  =  A  +  B  (x  -  m)  +  C  (x  -  m)2  +  D  (x  -  m)3, 
etc (a) 

Differentiate  (a)  successively: 

/'(*)    ==  B  +  2  C  (x  -  m)  +  3  D  (x  -  m)2 

+  4  E  (x  -  mf  + etc. 

/"O)  =  2  C  +  6  D  (x  -  m)  +  12  E  (x  -  m)2  + 

/"'(#)  =  6  D  +  24  E  (x  -  m)  + 

/"(*)  =  24  E  + 

Since  x  is  assumed  to  have  any  value,  let 

x  =  m. 

Then  f(m)  =  A  or 

f'(m)   =  B, 


/"(»)  =  ^  C, 


Elementary  Calculus.  259 


Substituting  in  (a) 
•(x)  =  l(m)+t'(m)(x-m) 


Example:  Develop  log  x  in  powers  of  (x  —  2). 
/    (*)=  log*,  /  (2)  =  log  2. 


fiv(x)  =  -   -4,  /w(2)  =  -  I,  etc. 

Hence  log  x  =  log  2  +  i  (x  -  2)  -  i  (x  -  2)* 

+  i  (x  -  2)3  -  |  (x  -  2)4  +  .  .  .  .  . 

ART.  53.  If  in  formula  (b),  m  be  made  o,  which  is 
clearly  permissible,  since  no  restrictions  were  placed  on  its 
value,  the  formula  becomes  the  development  for  f(x)  in 
terms  of  x: 


+  .  .  .  .  .  (b) 

where  /(o),  /'(o),  etc.,  mean  the  values  of  /(#),  /'(#),  etc., 
when  #  is  replaced  by  o. 

Example  :  Develop  cos  x  in  terms  of  x. 

I  (x)  =  cos  x,  I  (o)   =  cos  o  =  i. 

/'  (x)  =  —  sin  x,  /'  (o)  =  -  sin  o  =  o. 

f'(x)  =  -  cos*,  /"(o)  =  -  coso  =  -  i. 


260  Elementary  Calculus. 

/"'(#)  =  sin  x,  /"'(o)  =  sin  o  =  o. 

/Iv  (x)  =  cos  x,  /IV(°)  =  cos  o  =  i,  etc. 

Substituting  in  (b): 


which  is  the  expression  from  which  cos  x  is  computed. 
For  example,  to  find  cos  30°  =  cos  [—  rad.  J , 

f-T  (-V  f-V 

cos.  30°=  «  _  IS/.  +  JfiZ.  _  &L  +  etc.     (W  =  3.1416.) 

24          720 


.00313 


24 24 


1.00313 

—  (  —  1  =  —  .0000' 
—  .13711 

cos  30°  =        .86602  approx.     720 

-  .13711 

The  series  (b)  (and  its  special  form  bt)  is  known  as 
Maclaurin's  Series  from  its  discoverer. 

ART.  54.  It  is  frequently  necessary  to  express  a  func- 
tion of  two  quantities  in  the  form  of  a  series  of  powers  of 
one  of  them,  as  for  example,  }(h  +  x)  in  powers  of  x. 

The  process  is  entirely  analogous  to  that  employed  in 
the  development  of  Maclaurin's  Series,  and  the  result  is 
known  as  Taylor's  formula. 

Assuming  that  )(h  +  x)  can  be  developed  in  powers  of 
xt  and  regarding  h  as  constant: 

Let     }(h  +  x)  =  A  +  Bx  +  Cx2  +  Dx3  +  Ex4  +      (c) 


Elementary  Calculus.  261 

Taking  the  derivatives  with  respect  to  x, 

j'(x    +h)=        B  +    2Cx+    3  Dx2  +  4  Ex3  -f  .  .  . 

j"(x  +h)=    2  C  +   6  D*  +  12  E*2  +  .  .  . 

/"'(*  +  A)=:    6D  +  24  E*  +  .  .  . 

}™(x  +  h)  =  24  E  +  .  .  . 

Since  this  series  must  be  true  for  all  values  of  x,  being  an 
identity,  it  is  true  when  x  =  o;  hence  setting  x  =  o  in  this 
series  of  equations  we  are  enabled  to  determine  the  con- 
stants, thus: 


(2  =  2  X  i=  Z2) 
(6-3X2  Xi=  Zs). 


Substituting  in  (c) 
f(x  +h)  =  j(h}  +  f'(h)  x 


Z2 


Where  f(h\  j'(h\  etc.,  mean  the  values  of  f(x+h),  /'(x+h), 
etc.,  when  x  =  o. 

ART.  55.  It  will  be  evident  upon  consideration,  that  the 
binomial  theorem  as  encountered  in  algebra  is  a  special 
form  of  Taylor's  formula.  The  utility  of  these  develop- 
ments of  Maclaurin  and  Taylor,  depends  upon  the  rapidity 
with  which  they  converge. 


262  Elementary  Calculus. 

As  the  series  developed  by  these  two  formulae  is  usually 
infinite,  there  is  always  a  residual  error  in  taking  the  sum 
of  a  limited  number  of  terms  as  the  value  of  the  function 
thus  expanded.  A  discussion  of  this  error  is  unnecessary 
here;  it  will  be  sufficient  for  us  now  to  observe  that  a 
series  has  satisfactory  convergence,  if  the  successive  terms 
decrease  rapidly  in  value,  and  after  a  limited  number  of 
terms,  approach  zero. 

It  is  usually  an  effective  test  of  convergence,  when  the 
nth  term  of  a  series  can  be  readily  expressed,  to  find  the 
ratio  between  the  (n  +  iYh  and  nth  terms.  If  this  ratio 
approaches  zero  as  n  approaches  infinity,  the  series  is  con- 
vergent, otherwise  divergent,  and  hence,  useless  for  prac- 
tical purposes. 

Example  :  To  test  convergency  of  sine-series. 

T3  V5  T7 

sin  #=#-—+- —  +    ..   . 

Z3  /5  /7 


Inspection  of  the  relation  between  the  coefficients  of  x, 
the  denominators,  and  the  corresponding  term  number, 
gives  the  nth  term  as  above.  The  (n  +  i)tk  term  like- 
wise is, 


Z_2  n  +  i 
If  then  the  value  approached  by  the  ratio, 


.  n 


X2n~ 


,       .   ~   .. 
as  n  approaches  infinity, 


/_%  n  —  i 
is  zero,  the  series  is  convergent,  otherwise  not. 


Elementary  Calculus.  263 


(2  n  +  i)  2  n 


oifn-oo. 


Z_2  n  —  i 
Hence  the  sine-series  is  convergent. 

It  is  to  be  observed  that  it  is  only  the  absolute  values  of 
the  terms  that  are  considered,  as  the  sign  does  not  affect 
the  ratio.  There  are  numerous  more  complicated  tests 
for  convergency,  but  they  do  not  come  within  the  scope  of 
this  book. 

EXERCISE  VIII. 

1.  y  =  4  x3  —  8  x2  +  2  x  —  i,  find  — ^  • 

2.  y  =  x3,  find  —2-  • 

dxr 


4.   y  =  x  log  x,  find  —~  • 

(LOO 


5.  y  =  log  (ex  +  er*\  find  -^  • 

dxr 

6.  y  =  &  (x2  -  4  x  +  8),  find  ^ 

dx3 

T        ,.       ,     d4V 

7-  y  = 

8.   7  = 


10.   v  =  log  sin  ap,  find  — ^  • 


264  Elementary  Calculus. 

11.  y  =  sin  2  x,  find  —*•• 

12.  y  = ,  find  —2. 

i  —  #  efar 

13.  ^  =  e2*  (V  -  2  #  +  i),  find  -^2- 

14.  ^y  =  eax,  find 

id.   -y  =  e^sin  #.  show  -^ 2  ^~ 

dx2  dx 

1 6    y  =    - — i— —  ,  express    — ^  in  terms  of  y. 

ex  _  ^-x  ^2 

17.  -y  =  ^2ez,  show  that  —^=  6  ex(x  -{-  i)  +  y. 

n  /\^ 

18.  »=!  +  *«•,  find  g. 

19.  *?  -  3  a*y  + /,  find  j-2. 

20.  &2^c2  —  a2  y2  =  a2b2, 

21.  y2  =  2  £#,  find  -^ 

9     /•    i    d3y 

22.  jcy  =  tr,    nnd    •-*• 

^2. 

23.  e*4*  =  ^y,  find 


dx 


JH,  find  -     in  terms  of  y  and  a. 


25.   y5  =  a2  #,  find 


—    . 


26.   rv  =  r  vers-1  ^  -  \A  ^  -  /,    find    -       in  terms 
/•  ax 

of     and  r. 


Elementary  Calculus.  265 

EXERCISE  IX. 

Expand  by  Maclaurin's  formula: 
i.   sin  x         (in  powers  of  x\ 


2. 


3.   log*        (in  powers  of  (x  -  i)) 

4-  "  (in  powers  of  x}. 

i  —  x 

5.  ex  (in  powers  of  (x  —  2)). 

6.  -  (in  powers  of  (x  —  h)). 

Expand  by  Taylor's  formula  in  powers  of  x: 

7.  sin  (n  +  x).  10.   log  sin  (h  +  x) 

8.  \A—  ^2-  ii.   sec  (a  +  x). 

9.  ea+x.  12.    (a  -  x)n. 


CHAPTER   VI. 
EVOLUTION    OF    INDETERMINATE    FORMS. 

ART.  56.     Functions  of  a  variable  which  reduce  to  such 

forms  as  —  ,    -  —  ,  o,  oo  ,  etc.,  for  certain  values  of  the  vari- 

o      oo 

able  are  called  indeterminate,  because  we  are  unable  to  divide 
o  by  o,  or  oo  by  oo  directly,  but  must  approach  the  quotients 
by  a  circuitous  path. 

The  consideration  of  a  definite  example  may  make  the 
idea  clearer 

oc5  —  i 

wen  x  =  i. 


o<?  —  8        o       , 

and  —  =  —  when  x  =  2, 

x  —  2        o 

2  x  —  oc2  —  i        o      , 

and  -  =  -  when  x  =  i. 

3  ar  —  2  #  —  i       o 

Evidently  —  does  not  mean  the  same  thing  in  all  these 
o 

cases,  nor  in  the  multitude  of  similar  cases  that  might  be 
cited.     Having  practically  an  infinite  number  of  possible 

values  then,  the  expression   -     is    indeterminate.     It    will 

o 

be  recalled  that  in  '  discussing  the  differential  quotient,  it 

266 


Elementary  Calculus.  267 

was  remarked  that  although  two  quantities  may  each  be 
too  small  (or  too  large)  for  individual  comprehension, 
they  might  yet  have  a  finite,  readily  expressible  ratio,  if 
they  belonged  to  the  same  order  of  smallness  (or  largeness). 
To  use  a  somewhat  inadequate  illustration,  two  typhoid 
bacilli,  though  each  hopelessly  beyond  the  reach  of  our 
ordinary  senses,  could  be  readily  compared  with  one  another 
and  their  relative  size  could  be  expressed  by  a  very  simple 
number.  Although  a  bacillus  is  not  infinitely  small,  the 
same  illustration  may  be  extended  indefinitely.  As  the 
chemist  has  to  approach  the  problem  of  his  inconceivably 
small  atom  and  the  astronomer  of  his  inconceivably  vast 
distances,  indirectly,  so  we  will  have  to  deal  with  our  zeroes 
and  infinities. 


—  i 


To  return  to  the  expression 

x—  i 

Before  giving  x  any  definite  value,  divide  the  numerator 

x5  —  i 
by  the  denominator,  then  -       —  =  x4  -}-  x3  -}-  x2  -{-  x4-  i. 

x  —  i 

If  in  this  expression  we  give  x  a  constantly  decreasing 
value  >i,  the  integral  function  will  clearly  approach  more 
and  more  nearly  the  value  5,  while  the  fraction  approaches 

the  value  —  .     It  is  easy  to  infer  then  that  when  x  is  actually 

i,  the  value  of   --  becomes  exactly  5. 
o 

Again  the  expression 

2  x  —  x2  —  i 


3  x2  —  2  x  -i 

may  be  shown  to  approach  --  J  as  x  approaches  oo, 
if  we  first  divide  both  numerator  and  denominator  by 
x2. 


268 


Elementary  Calculus. 


ART.  57.     To  find  a  general  method  for  evaluating  an 
indeterminate. 


Let 


By  Maclaurin's  formula, 


—  when  x  =  a. 
o 


/  \       if  \   i    it f  \  f  \    i      /    \a )  ,  so 

(X)  =  f(a)  +  j'(a)  (x  -  a)  +   -L±-L  (x  -  a)2 


'(a)  (*  -  a) 


But  /(a)  =  o  and  <j>  (a)  =  o  by  hypothesis. 
.   /(*) 


If 


(dividing  numerator  and  denominator  by  x  —  a) 
(since  (x  —  a),  (x—  a)2,  etc.  =  o  when  x  =  a). 

' ,     '        still  equals  —  for  x  —  a, 


it  is  clear  that  the  expression  reduces  to   -*— - — '- ,    if 


Elementary  Calculus.  269 

<f>'(x)  are  replaced  by  their  values,  o,  and  numerator  and 
denominator  be  again  divided  by  x  —  a. 

Hence  when  1&  =  -   for  x  =  a, 

<*        o 


etc 

<j>(X)       </>'(*)       #'(*)' 

A  rule  may  be  stated  thus  : 

Take  the  successive  derivatives  of  numerator  and  denom- 
inator (as  distinct  junctions)  until  a  derivative  is  found, 
say  fn(x),  which  is  not  zero  for  x  —  a.  Then, 


Ll       T-.     , 
Example  :  Evaluate 


is  the  value  sought. 

X 

tan  x  —  sin  x  cos  x        o 


, 
x3  o 

when  x  —  o. 

tan  x  —  sin  x  cos  x        j(x} 

-?-       =^)' 

tan  x  —  sin  x  cos  x  _  sec2  x  —  cos  2  #  +  sin  2  # 

"^^  3^ 

(taking  derivatives). 

This  expression  corresponding  to     '  ^    •  still  equals  _  • 

<p  (x)  o 

Hence  taking  second  derivative, 

tan  x  —  sin  x  cos  x    _  sec2  a;  —  cos2  x  +  sin2  x 

x3  3^ 

2  sec  x  tan  #  +  2  cos  x  sin  #  +  2  sin  jc  cos  Jg 

6x 
sec  5P  tan  ^  +  2  sin  #  cos  x 


(collecting  and  dividing  by  2). 


270  Elementary  Calculus. 

ifrr  /  \* 
This  is  still  -  •       Taking  third  derivative    '  fff{  } 

sec3  x  +  sec  x  tan2  x  +  2  cos2  x  —  2  sin  2x  _  3_ 

3  3 

=  i,  when  x  =  o. 

.-.  tan*  -sin*  cos*    =       whfin  x  _  Q 

tf3 

ART.  t;8.     If   -"—2  =    —   when  x  —  a,  a  simple  trans- 

0(x)          co 

formation  reduces  the  expression  to  the  form  —  ;  for 

o 


O      r 

=  —  for 


<j)(x)  i 


If  j(x)  =  o  and  <j)(x)  =  cc   for  ^  =  a, 

then  /(#)  .  (f>(x)  =  o  .  oo.  an  indeterminate, 

o 


but  fix)  .  <j>(x)  = 

I  O 


By  using  the  logarithms  of  the  functions  as  an  interme- 
diate   step,    expressions  like    i°°,    o°°,    00°,    etc.,    may    be 

reduced  likewise  to  —  .     For   example,  let   f(x)  =  i  and 
o 

<p(x)  =  oo,     when  x  =  a. 

Then  }(x)]^  =  ix. 

Let  y  =  [/(*)]*<*>. 

Taking  the  log  of  both  sides : 


Log  y=(f)(x)  log  /(*)  =  =  o   when  »  =  a. 

i  o 


Elementary  Calculus.  271 

In  these  cases  we  get  eventually  the  logarithm  of  the 
function,  from  which  the  function  itself  is  readily  found. 

/  x  \  tan  -^2- 

Example :  Evaluate  f  2  -    -  j      2  «  ,  when  x  =  a, 

T\tan  ^—r 

1  =  i    ,  when  x  =  a. 


- 5)- 

log  (2  -  £ 

Then  log  y  =  tan  ^   log  ( 2  -  -  )= * 

2  a      &  V        0  /  .  WP  - 


o_ 
cot  —  ° 

2 


-   I 


/.     log  y  = 


2  a  2  a          2  a 

i  i 


2  a  —  je  a         2       , 

=  — •  =  — ,  when  x  =  a. 


JL  esc2  —         - 
2  a          2  a       2a 


I  ft  \  tan  ff^  2 

That  is,  log  y  =  log  (2 )      2  a  =  -,  when  ^  =  a. 

\        a  /  TT 


tan  E*_  2 

(a  —  a)      2  a  =  e  -IT 


Example :  Evaluate  (a x  —  i )  x,  when  #  =  oo . 

i  a. 

(a*    -  i)  x=  (a00    -  i)  oo  =  (a°  —  i)  oo  =  o.oo, 

when  x  =  GO  . 


272  Elementary  Calculus. 

But  a7  -i*  =  £llLl..  o 


1  ** 

X  I_ 

when  x  =  oo . 

EXERCISE  X. 

Evaluate : 

x.  J2£Z|Wheny=i. 
y-  i 

e*  -  g-* 

2.   ,  when  x  =  o. 

tan  x 

2.   4  ^  sin  jg 2_7r     wnen  ^_-_, 

cos  x  2 


4-   -^  ---         , 
cos2  #      i  —  sin  6 


5.  tf*"1,  when  #=  i. 


6.  'sin  y)tan  y  ,  when  3;  =  -. 

2 

gZ       I        g—  2    _     2 

7.  —  —  —  -  -  ,  when  z=o. 

2 

8.  (i  +  ^)"-i 


.    [  i  +  -  )    ,  when 
V         */ 


x  =  oo  . 


Elementary  Calculus. 


273 


sin     x        , 

10.   --  :  —  ,  when  x  =  o. 
tan-1  x 


ey  sin  y  —  y  —  y         , 
"•  —  —  *:  i  when 


log  sin  2  *       , 
12.   -  —  ,  when  oc  =  o. 

log  sin  x 


13-    (m*  —  i)*,  when#  =  oo. 

14.  — —  ,  when  x  =  i. 

£>X     p  /Y*     T 

O  O  vV  X 

15.  -      -  —  rx—  ,  when  *  =  i. 
log*      log* 

16.  (cos  26)°  ,  when  6=0. 

17.  (log  *)a;-1,  when  *  =  i. 


18.     °^  ^  ,  when  *  =  o. 
esc  * 

19     (i  —  tan  *)  sec  2  x,  when  *  =  - 

4 

20.  c~x  log  *,  when  *  =  oo . 

21.  [log  (e  +  z)]*,  when  2=0. 


22 


(*\  ^* 

2  —  —  ]  tan  -  — ,  when  x  =  n. 
n)          2n 


sec 


24. 


log  (i  -  x) 


,  when  *  —  i . 


274  Elementary  Calculus. 


25.  —  —  cot  x,  when  x  =  o. 

00 

26.  *  -cos       when  ^  =  Q 


x  —  sin-  jc       , 
27.   -  —  —  —  ,  when  x  =  o. 


28.    2X  sin  —  ,  when  x=  oo  . 


29.  (sin  #)sin  *,  when  ^  =  o. 

i 

30.  rv  e*,  when  ^c  =  o. 

5C2  +   2  COS  3f  —  2          , 

31.  -  -  ,  when#  =  o. 


i  —  sin  x  +  cos  #1  TT 

32.   —  —  —  ,  when  jf  =   — 

sin  5f  +  cos  x  —  i  2 


33- 


34-  -,  when  3^=0. 

(ey-  i)? 

35.   rv  tan  ^  —  —  sec  ^,  when  x  =  - 
2  2 


—  )   ,  when  0  =  oo . 


CHAPTER  VII. 
MAXIMA    AND    MINIMA. 

ART.  59.  When  a  function  has  a  maximum  value  it  is 
an  increasing  function  until  it  reaches  the  value  then  a 
decreasing  function  just  afterward,  otherwise  this  value 
would  not  be  a  maximum.  Since  the  derivative  of  a  func- 
tion is  the  ratio  between  its  increase  and  the  increase  of  its 
independent  variable,  if  the  function  is  increasing  with  the 
variable  the  derivative  will  be  positive;  if  it  is  decreasing 
as  the  variable  increases  the  derivative  will  be  negative. 
Hence  when  a  function  passes  through  a  maximum  value 
its  derivative  changes  from  positive  to  negative,  and  in 
order  to  do  this  it  must  pass  through  the  value  zero,  if  it  is 
continuous.  A  similar  process  of  reasoning  shows  that 
when  a  function  passes  through  a  minimum  value  the  deri- 
vative also  passes  through  zero  from  negative  to  positive. 
It  is  to  be  remembered  that  since  a  function  depends  upon 
its  variable  for  its  value,  it  can  be  made  to  take  any  number 
of  values,  as  near  together  as  we  please,  by  giving  the 
variable  a  suitable  series  of  values,  that  is  provided  always 
that  the  function  is  continuous. 

A  graphic  illustration  may  make  this  plainer. 

Since  in  general  any  function  may  be  represented  graph- 
ically by  a  curve,  let  the  curve  AB,  Fig.  23,  represent 

y  =  /(*)• 

Since  the  derivative  of  a  function,  represented  by  a 
curve,  is  the  slope  of  its  tangent  at  any  given  point,  the 
change  of  the  derivative  and  the  tangent  slope  are  synony- 

275 


276 


Elementary  Calculus. 


mous.  Suppose  T  is  a  maximum  point  for  the  value 
x  =  OD.  A  glance  at  the  figure  will  show  that  starting, 
say  with  the  tangent  MN  at  A,  the  slope  of  this  tangent  as 
the  point  of  tangency  moves  from  A  to  T  will  be  constantly 
positive  (the  inclination  being  an  acute  angle,  as  AMO)  but 
constantly  decreasing;  at  T  the  slope  will  be  zero,  for  the 
tangent,  RS,  is  parallel  to  the  jc-axis;  beyond  the  point  T, 
the  inclination  of  the  tangent  is  an  obtuse  angle  as]  PQ#, 
and  hence  its  tangent  is  negative,  but  it  will  still  decrease 


Fig.  23. 

in  general.  Therefore,  as  indicated,  the  derivative  of  the 
function  which  is  always  equal  to  these  slopes,  will  pass 
from  positive  to  negative  through  zero.  But  a  function 
may  pass  through  zero  or  infinity  without  changing  its 
sign,  so  even  when  the  derivative  is  zero  there  may  not  be  a 
maximum  or  minimum.  Hence  it  is  necessary  to  deter- 
mine in  a  given  case  whether  a  maximum  or  minimum 
exists. 

Recall  the  fact  cited  above,  that  the  slope  decreases  to 
zero  before  a  maximum  and  continues  to  decrease  (because 
it  is  negative)  after  a  maximum,  hence  the  derivative  is  a 


Elementary  Calculus.  277 

decreasing  function  at  a  maximum,  hence  its  derivative, 
that  is,  the  second  derivative  of  the  original  function,  will 
be  negative  from  our  definition  of  a  derivative. 

An  examination  of  the  figure  around  the  point  F  (a 
minimum)  will  show  that  at  a  minimum  the  slope,  and 
hence  the  derivative,  passing  from  negative  to  positive 
through  zero,  is  an  increasing  function,  hence  its  deriva- 
tive, that  is,  the  second  derivative  of  the  function,  is 
positive.  This  suggests  a  general  method  for  determining 
maxima  and  minima,  as  follows  : 

Since  the  first  derivative  is  always  zero  at  a  maximum  or 
minimum  point,  if  the  first  derivative  is  found  and  set 
equal  to  zero,  the  value  of  the  variable  found  from  this 
equation  will,  in  general,  be  one  of  the  co-ordinates  (usually 
the  abscissa)  of  the  maximum  or  minimum  point  on  the 
curve  representing  the  function.  To  determine  whether 
it  is  a  maximum  or  minimum,  the  second  derivative  is 
found,  and  if  it  is  negative  in  value  for  this  value  of  the 
variable,  the  point  is  a  maximum;  if  positive,  it  is  a  minimum. 

ART.  60.  It  may  happen  that  the  second  derivative  is 
also  zero  for  this  value  of  the  variable,  and  hence  indeter- 
minate as  to  sign.  In  this  case  it  is  clearly  desirable  to 
expand  the  function  in  the  neighborhood  of  this  value  of 
the  variable  that  its  character  may  be  more  readily  seen. 

If  (fx)  is  the  function,  and  x  =  a  be  the  value  found  from 
f(x)  =  o,  then  f(a  —  h)  and  f(a  +  h)  will  represent  the 
value  of  the  function  immediately  before  and  immediately 
after,  respectively,  its  value  for  x  =  a,  h  being  a  quantity 
which  can  be  made  as  small  as  desired. 

By  Taylor's  formula : 


278  Elementary  Calculus. 

j(x  -  h)  =  j(x}  -  f(x)h  +  £^)  h2  -  £^U3  +  . 

Z2  Z3 

Replacing  x  by  the  value  a,  and  transposing  /(a), 
f(a+h)-f(a)  =  f(a)h 


f(a  _/,)  _  j(a)  »  -/ 

Z2 

Now  since  h  is  to  be  taken  exceedingly  small,  its  square, 
cube,  etc.,  in  the  developments  will  be  insignificant,  and 
hence  the  values  of  the  above  expressions  will  practically 
equal  the  first  terms  of  their  development.  That  is, 
f(a  +  h)  —  j(a)  will  have  the  same  sign  as  /'(#)/£,  and 
j (a  —  h}  -  /O)  will  have  the  sign  of  -  f(a)h.  But  if 
there  is  a  maximum  or  minimum  at  a,  f(a  -f  h)  and  }(a—h) 
must  have  the  same  value,  because  if  it  increases  to  a 
maximum  it  must  decrease  beyond  the  maximum,  and 
hence  have  the  same  value  just  before  and  just  after,  as 
the  sun  has  the  same  altitude  at  the  same  time  before  noon 
and  after,  noon  being  its  maximum  elevation. 

But  the  only  way  }'(a)h  and  —  f(a)h  could  both  have 
the  same  value  would  be,  that  both  equal  zero,  that  is,  that 
j'(a'}=  o  [/'(a)  being  value  of  j'(x)  when  x=  a],  which 
verifies  our  former  conclusion. 

If  /'(a)  =  o,  then, 


and 


Z2  Zs 

Since  h  is  so  small,  h2  is  much   larger  than  h*  or  any 
higher  power,  hence  j(a  +  h)  —  f(a)  and  }(a  —  h)  —  f(a) 


Elementary  Calculus.  279 

are  determined  by  '    ^a'  h2,   and  hence  are  positive  if  I"  (a] 
is  positive,  and  negative  if  f'(a)  is  negative 

for  /"  (a)  determines  the  sign  cf  the  term  L^LL  h?\. 

But,  when  f(a  +  h)  —  f(a)  and  }(a  -  h)  -  f(a)  are 
both  negative,  f(a)  is  a  maximum,  since  it  is  greater 
than  the  values  on  either  side  of  it  \j(a  -f-  h}  and  f(a  —  h)]; 
likewise,  when  they  are  both  positive,  f(a)  is  a  minimum. 
But  these  conditions  prevail,  respectively,  when  f"  (a)  is 
negative  and  when  /"(#)  is  positive,  which  verifies  our 
second  conclusion  above. 

If  f"(a)  is  also  zero,  then, 


and 


A  course  of  reasoning  exactly  as  before,  will  show  that 
for  a  turning  value  (maximum  or  minimum) 

^M  h3  and  -  f"(a)     h3  must  equal  zero, 


that  is,  /'"O)  =  o, 

and  when  fiv(a)  is  positive  there  is  a  minimum;  when  /iv(a) 
is  negative  there  is  a  maximum,  etc. 

Hence  the  rule  : 

A  function  has  a  maximum  or  minimum  value  at  x  =a, 
if  any  number  of  the  successive  derivatives,  beginning  with 
the  first,  is  zero  for  x  =  a,  provided  the  first  that  does  not 
equal  zero  is  of  even  order,  being  negative  for  a  maximum 
and  positive  for  a  minimum. 


280  Elementary  Calculus. 

The  values  of  the  variable  which  cause  the  first  deriva- 
tives of  a  function  to  vanish  are  called  critical  values. 
Example:  Find  turning  values  of  (x  —  i)3(x  —  2)2. 

/(*)=    (*-l)3(*-2)2 
/'(*)  -    3(*  -    l)2(*  -    2)2  +    2(X  ~    l)3(*-2) 

whence  (x  —  i)2(x  —  2)  (5^  —  8)=  o, 

x=  i,  i,  2,  f. 
/"(*)  =  2  (x  -  i)  (x  -  2)  (5  x  -  8)  -f  (*  -  i)2(5  *  -  8) 

-f  5(*-   l)2(*~2). 

When  x=  i, /"(*)  =  o. 

#  =  2,  /"(#)  =  2  (positive). 
*=  §,/"(*)  =  -M    (negative). 
Hence  for     x  =  2,  there  is  a  minimum, 
and  for  x  =  f ,  there  is  a  maximum. 

Since  /"(#)  =  o  for  x  —  i,  it  is  necessary  to  find  the 
third  and  fourth  derivatives. 

j'"(x)  =  2  (30  x2  —  84  x  +  57)  =  6  when  x  —  i. 

Hence  there  is  neither  maximum  nor  minimum  at  x  =  i. 

Example  :  What  are  the  dimensions  of  the  cylindrical 
vessel  of  largest  contents  that  can  be  made  from  3234 
squire  inches  of  tin  plate,  not  counting  waste? 

Since  3234  square  inches  will  constitute  the  surface  of  the 
cylinder  (one  base)  when  completed, 

2  nrh  +  xr2  =  3234      ......     (i) 

Volume  =  nr*h (2) 

which  is  to  be  a  maximum. 


From  (i)  h  =  • 


2  nr  2r 


=  2l1 

7  J 


Elementary  Calculus.  281 

Substituting  in  (2) 


Since  a  constant  does  not  change  value  it  cannot  affect 
a  maximum  or  minimum,  hence  any  constant  factor  may 
be  ignored,  in  searching  for  turning  values. 

Say  then,  /  (r)  =  1029  r  —  r*, 

l'(r)=  1029    -  3r2=  o, 
whence  r2  =  343,  r  —  7  v  7. 

f'(r)  —  —  6  r  which  is  negative,  hence  r  =  7  V/  gives  a 
maximum. 

From  (i)  h=  7  X/7  for  r  =  7  x/y.  Hence  the  cylinder 
will  have  greatest  contents  when  its  radius  equals  its 
altitude. 

EXERCISE  XI. 

Find  maxima  or  minima: 

y-  8  (z  +  9)  (z  -  2) 

~~~ 


3  . 

O  /  \  o 


i  +  x       i  —  x 
^2  +  2  M  +  3 


/  +  y  -  i  w2  +  i 

7.  Divide  a  line  i'  long  into  two  parts,  such  that  their 
product  will  be  a  maximum. 

8.  Find  the  greatest  rectangle  that  can  be  inscribed  in 
a  circle  of  radius  6". 

9.  Find  the  volume  of  the  greatest  cylinder  inscribed  in 
a  sphere  of  8"  radius. 

10.  Find  the  greatest  cone  in  the  same  sphere. 

n.  Show  that  it  takes  the  least  amount  of  sheet  iron  to 
make  a  cylindrical  tank  closed  at  both  ends,  when  its 
diameter  equals  its  height. 


282  Elementary  Calculus, 

12.  Find  the  greatest  cylinder  that  can  be  inscribed  in 
a  right  cone  of  radius,  r,  and  height,  h. 

13.  Calling  the  E.M.F.  of  a  cell,  E;  internal  resistance  r, 

jr 

external  resistance,  R,  and  current,  C,  C  =  -      —  and  the 

r  -f  R 

power,  P  =  RC2.     What  value  of  R  will  make  P  a  maxi- 
mum? 

14.  Find  the  shortest  straight  line  that  can  be  drawn 
through  a  given  point  (m,  n)  and  limited  by  the  axes. 


CHAPTER   VIII. 
PARTIAL    DERIVATIVES. 

ART.  61.  Up  to  this  time  functions  of  one  independent 
variable  only  have  been  considered,  but  an  expression 
may  be  a  function  of  two  or  more  independent  variables. 
A  function  of  two  variables,  x  and  y  say,  is  symbolized 
thus: 

/  (x,  y),  <f>  (x,  y),  F(x,  y),  etc. 

Continuous  functions  only  give  important  general  results, 
and  a  function  of  two  variables  is  continuous  about  any 
specific  values  of  these  variables,  say  x  =  h,  y  =  k,  when 
the  function  runs  through  an  unbroken  series  of  values  (as 
near  together  as  we  please)  as  its  variables  run  through 
corresponding  series  of  consecutive  values,  in  the  vicinity 
of  h  and  k. 

ART.  62.  The  derivative  of  a  function  of  two  (or  more) 
variables  found  by  considering  all  the  variables  except  one, 
as  constants,  is  called  its  partial  derivative  with  respect  to 
the  variable  that  changes.  For  example,  4  xy  +  3  y2  is 
the  partial  derivative  with  respect  to  x  of  the  function 
2  x2y  -f  3  xy2  +  y3  (regarding  y  as  a  constant)  and  is 
represented  thus: 

—  (2  x2y  +  3  xy*  +  /)  =  4  xy  +  3  y2. 


Ifs=  2X2y  +  3xy2  +  y>,then=4*y+3y2     •     (i) 


Likewise  the  partial  differential,  with  respect  to  x,  is  repre- 
sented thus: 

283 


284  Elementary  Calculus. 

'dyZ  =  4  xy  dx  +  3  y2  dx (2) 

Evidently  9xz  =  — -  dx.  since  (2)  equals  (i)  multiplied  by  dx. 

'dx 

Similarly,  dyz  =  (2  x2  +  6  xy  +  3  y2)  dy (3) 

By  the  principles  of  differentiation  already  known, 

dz  =  4  xy  dx  +  2  x2  dy  +  3  y2  dx  +  6  xy  dy  +  3y2  dy.    (4) 

A  comparison  of  (2),  (3)  and  (4)  will  show  that 

-^         .     --N  OZ     i  OZ    , 


That  is,  in  this  case  the  total  differential  equals  the  sum 
of  the  partial  differentials. 

In  Art.  4,  and  succeeding  articles,  it  was  explained  that 
a  differential  quotient  (or  derivative)  was  the  ratio  of  the 
increase  of  a  function  to  the  increase  of  its  variable  when 
these  increments  were  indefinitely  small.  This  may  be 
expressed  thus:  if  y  =  f(x), 


dy  _ 


hesa 


dx  Ax 

Likewise  in  a  function  of  two  variables,  x  and  y  say,  if 

z  - 


as 


[(  =  )  is  a  symbol  meaning  "  approaches.  "] 


Also         »  =      *.yy-.yasA    ^^ 

oy  A;y 

in  the  first  case  y  remaining  constant  while  x  changes  to 
x  +  A#,  and  in  the  second  #  remaining  constant  while  y 
changes  to  y  +  A^. 


Elementary  Calculus.  285 

Now  let  these  changes  take  place  together  in  the  same 
function  and  we  have, 

3+  Az  =  f(x+  A*,  y  +  Ay)       .     .     .(a) 
But  the  result  would  plainly  be  the  same,   if  instead  of 

changing  simultaneously,  x  should  change  while  y  remained 

constant  and  then  y  would  change  while  x  +  A#  remained 

constant. 

From  (a),  Az  =  /  (x  +  A*,  y  +  Ay)  -  }  (x,  y), 

or  changing  successively, 

Az  =  /  (x  +  A*,  y)  -  j  (x,  y) 

+  /  (x  +  A*,  y  +  Ay)  -  /  (x  +  A*,  y). 
Az  =  /(*+  A*,y)-/(*,y) 
A#  A# 

.  /  (x  +  kx,  y  +  Ay)  -  /  (x  +  AJC,  y)      A^ 
Ay  '  A* 

(Multiplying  and  dividing  the  last  two  terms  by  Ay,  and 
dividing  through  by  A*.)    By  definition  of  derivative, 

/(*+  A*-  ?)  -/<*'?)  [as  A*  =o]-|L. 

A#  o# 

and 

,      ,     ^  0]  =  Jz  . 


Ay  3y 

»T.I    A  •         dz       'dz    .   'dz   dy       j        ~dz   -,    .   3z   , 
That  is,      —  =  -  --  \-  —  -f-oidz=  —dx+  —dy. 
dx        ox       oy  dx  ox  oy    ' 

Hence  the  result  found  in  the  specific  example  above  is 
shown  to  be  general  for  all  continuous  functions,  namely: 
The  total  differential  equals  the  sum  of  the  partial  differen- 
tials, each  being  multiplied  by  the  differential  oj  its  inde- 
pendent variable. 

This  rule  could  be  easily  inferred  from  the  rules  already 


286  Elementary  Calculus. 

enunciated  for  the  differentiation  of  specific  forms  as,  for 
example,  the  product  of  two  or  more  variables,  wherein 
the  differential  is  found  by  regarding  all  the  variables  but 
one  successively  as  constant,  and  taking  the  sum  of  the 
results. 

ART.  63.  In  implicit  functions,  which  are  presented 
most  frequently  for  partial  differentiation,  the  form  is 

/  (*,  y)  =  o. 

An  implicit  function,  it  will  be  remembered,  is  one  in 
which  the  variables  are  thrown  together  in  the  various  terms, 
and  the  function  is  not  solved  explicitly  for  any  one,  like 
3  x2y  -  xy  +  7  xyz,  etc. 

From  our  rule, 


,  or  shortly,^- -?£. 
dx          3 


The  same  process  applies  to  any  number  of  variables, 
for  example,  if 

w=  <f>(x,  y,  z), 


dw  =  ^dx+  ^-dy  +  -~dz,  etc. 

ox  oy    '        oz 

ART.  64.     If  y  is  itself  a  function  of  x,  say  y  =  (f>(x), 
then  the  form 

dz  __  3z    ,    3z    dy 
dx        'dx       'dy  dx 


Elementary  Calculus.  287 

is  most  effective,  for  —2   can  be  found  from  y  =  d>  (x). 
dx 

Example  :  z  =  tan-1  2-2  and  x2  +  4  y2  =  i. 

x 

By  formula, 
dz 


.....  (a) 


From 


'dx 

3y 

dx 

—  2  y 

2 

X 

dj_ 

whence  -2-  = = —   since  y  =  — 

<fo          2  V  i  —  yf  4y  L  2      J 

Substituting  in  (a), 

dx  x2  +  4  y2      2y  (x2  +  4y2)  \2y  ) 

=  _   x  +  4^     _ E_  [since  ^  +  4  Vs  =  i]. 

2  y  2  y 

ART.  65.     Successive  partial  differentiation. 

A  function  of  two  or  more  variables  may  have  successive 
partial  derivatives  for  the  same  reason  that  was  given  for 
the  successive  total  differentiation  of  a  function  containing 
but  one  variable. 

The  process  is  indicated  thus: 


.      _         =          etc 
*' 


288  Elementary  Calculus. 

It  is  readily  shown  that 

.  ;  that  is,  the  order  is  immaterial. 


'dx'dy 


EXERCISE    XII. 


Find  SL  by  partial  derivatives: 
dx 

i.  a2y2  +  b2x2  =  a*b\ 
x3 


2  a  —  x 
4.   9  a/  =  x  (x  -  3  a)2. 

s.,-^5*.-5). 

6.   **  +  /  =  a*. 

1.  x=  r  vers-1  -  -  \/2  ry  -  y2. 
r 

8.z=  tan-1  y~  ;  show  that  x  —  +  y  —  =  o. 
9.  z  =  log  (tan  #  +  tan  ^  +  tan  w);  show  that 


sn  2 


+  sin  2  y  —  +  sin  2  M  -  -  =  2. 


ou 

10.  ^  +  /  -f  3  ^^  =  °;  find  "^  • 

11.  z  =  ^2y  +  ^Y2;  show  that 


Elementary  Calculus.  289 

12.  z=-       =J=      ^ ;  show  that  |^  +  |L; +|^  =  o. 

+  /  +  w»  a**    a/    3^2 


13.   z= 


14.   z  -  Vr^*2  -  /,   /  =  f8  -  ^2;  find 


CHAPTER   IX. 
DERIVATIVES    OF   ARCS,    AREAS,    VOLUMES,    ETC. 

ART.  66.  The  most  important  applications  of  the  deriva- 
tive have  to  do  with  curves  whose  equations  are  known. 
By  the  principle  of  minute  increments  the  characteristics 
of  a  curve  of  irregular  curvature  are  discovered. 

In  dealing  with  curves  it  will  be  helpful  to  regard  them 
as  described  by  a  point  moving  according  to  a  fixed  law, 
and  at  any  given  instant  having  the  direction  of  a  tangent 
line  to  the  curve  at  the  position  of  the  point  at  that  instant. 

Length  of  an  Arc. 

ART.  67.  Let  AB  be  an  arc  of  any  curve  (Fig.  24), 
P  and  Q  two  positions  of  the  describing  point,  6  and  (j)  the 


Fig.  24. 

angles  made  respectively  by  PQ,  and   the  tangent  at  P, 
MN,  with  the  #-axis,  to  find  the  length  of  the  arc  PQ. 

Draw  the  co-ordinates  of  P  and  Q,  (OT,  PT)  (OS,  QS). 

Then  TS  =  PR  =  A*  and  QR  =  Av. 

290 


Elementary  Calculus.  291 

In  the  right  triangle  PQR, 
chord  PQ2  =  PR2  +  QR2, 

that  is,  PQ2  =  A?  +  Ay2, 

or  PQ  »  V^+  ^V2- 

Dividing  by  Ax, 

•-.••« 


But  as  Ax  is  taken  smaller  and  smaller,   approaching 
zero,  the  chord  PQ  approaches  the  arc  PQ   (Q  moving 

down  toward  P),  and  eventually   -^  becomes  —  (where  s 

Ax  ax 

represents  the  arc). 


dx 
The  same  result  may  be  obtained  from  (b)  thus  : 

— ^  =  — —   .  — -^-    [multiplying  and  dividing  by  PQ]  ; 
Ax       PQ        Ax 


whence  T  +  .     .     .(from(b)) 

Ax      PQ 

But  — ^-eventually  equals  i,  since  the  chord  eventually 
equals  the  arc,  when, 

A£  _ds_ 

Ax  ~  dx 


Corollary :  The  tangent  MN  gives  the  ultimate  direction 
of  the  chord  PQ,  and  Ax  becomes  dx  and  Ay  becomes  dy 


29 2  Elementary  Calculus. 

at  the  same  time.      Since  by  what  has  been  said  in  Art.  n, 


from  (c) 
or 

Likewise, 


ds          , 

j^  =  V  i  +  tan2  <£  =  sec 


=  sn    . 


Volume  of  Solid  of  Revolution. 

ART.  68.    Let   the   arc   LN   revolve   about   the  #-axis, 
(Fig.  25)  to  find  the  volume  whose  surface  is  generated  by 


M 


Fig.  25. 

MN  =  As,  a  portion  of  LN.  This  volume  plainly  lies 
between  the  volumes  generated  by  the  rectangles  TNRQ 
and  MPRQ.  Since  these  will  be  cylinders,  calling  the 
volume  generated  by  MNRQ  (MN,  the  chord),  AV,  we 
have, 

TT  (y  +  A?)2 A*  >  AV>  ;ry2A# 

[x  =  OQ,y=  MQ,  A*  =  QR,  ky  =  NP]. 


Elementary  Calculus.  293 

Dividing  by  A#, 

TT  (y  +  Ay)2  >  AY  >  ^y2. 

As  the  arc  is  taken  shorter  and  shorter,  N  approaching 
M,  R  approaches  Q,  and  NR  approaches  the  value  MQ; 
that  is, 

y  +  Ay  approaches  y. 

But—  always  lies  between  n(y  +  Ay)2  and  Try2,  hence 
it  cannot  pass  Try2,  but  if  Tr(y  +  Ay)2  reaches  the  value 

of   Try2,    it   must    also   reach   it,   becoming   —  (generated 

dx 

by  the  arc). 


To  Find  the  Surface  Generated. 

ART.  69.  The  surface  generated  by  chord  MN  will  be 
that  of  a  cone-frustrum,  hence  calling  it  AS  (Fig.  25), 

AS=  Tr(2y  +  Ay)  MN. 

As  the  arc  is  taken  indefinitely  small,  N  approaching  M, 
the  chord  MN  approaches  its  arc  ds,  and  hence  AS 
approaches  dS,  the  surface  generated  by  the  arc,  as  Atf 
approaches  dx,  hence  finally  (dividing  through  by  Ax), 

^S  =  2       ds_  rgince  ^    =  Q  ag  N  approaches  M]. 
dx  dx 

But  -^  = 


CHAPTER  X. 
DIRECTION   OF   BENDING   AND    CURVATURE. 

ART.  70.  A  curve  is  said  to  be  concave  upward,  at  a  given 
point,  when  immediately  before  and  after  this  point  it  lies 
above  the  tangent  line  at  that  point. 

It  is  concave  downward  when  it  lies  below  the  tangent 
line. 

If  the  curvature  changes  concavity  at  a  point,  that  point 
is  called  a  point  of  inflection. 

In  Fig.  26  the  curve  is  concave  downward  at  A,  concave 


Fig.  26. 

upward  at  B,  and  has  a  point  of  inflection  at  C.  It  is 
evident  that  at  a  point  of  inflection  the  tangent  line  crosses 
the  curve. 

It  is  clear  also  that  the  conditions  for  downward  con- 
cavity are  the  same  as  for  a  maximum,  and  for  upward 
concavity  are  the  same  as  for  a  minimum. 

Since  the  second  derivative  is  negative  for  a  maximum 
and  positive  for  a  minimum,  at  a  point  of  inflexion  where 

294 


Elementary  Calculus.  295 

the  curve  changes  from  one  to  the  other,  the  second  deriva- 
tive must  change  from  positive  to  negative  or  vice  versa, 
that  is,  it  must  pass  through  zero  (or  infinity),  hence  solv- 
ing the  equation, 


gives  the  point  (or  points)  of  inflection  if  such  exist.  If 
f(x)  =  o  changes  sign  for  this  value  (or  these  values), 
there  is  a  point  of  inflexion. 

8  a3 

Example  :  Examine  y  =  -r-        -  for  inflexion. 
2 


/'(*)=- 


Substitute  in  /"(*),  x=  -^=  +  h  and  x  =    -^L  -  h 

\/3 


successively,  where  h  is  as  small  as  we  please. 
l6o'/4  a2  +4^+^-4 


Then    ?(*) 


,_   r  h   -h  4  » 

^3 


296  Elementary  Calculus. 


lA^ 
V  3 


++ 
V3 


and          /"(*)  = 


Since  &  is  so  small,  the  denominator  is  positive  in  both 
cases,  but  for  the  same  reason  ^-=  >  h2,  hence  the  second 


value  of  j"(x)  is  negative  and  the  first  positive,  and  hence 

x  =  ~^  \y  =   —     is   a  point  of   inflection,  as  is  also 
V3  L  2  J 

--  20  t3_a\    ky  tne  same  proof  t 


V  3 


CURVATURE. 


ART.  71.  If  two  curves  have  the  same  tangent  at  a 
point  of  intersection  they  are  said  to  have  contact  oj  the 
first  order:  that  is,  if  y  =  f(x)  and  y  =  F(#)  are  the  equa- 
tions of  the  curves,  then  for  a  point  of  intersection  the 
equations  are  simultaneous  and  we  may  combine  them 
any  way  we  please  to  find  />,  and 

t(P)=F(p)      ......     (i) 

Also  their  tangents  being  the  same, 
/'(/>)  -  Y'(p). 

[The  values  of  f(x)  and  Fx  (x)  when  x  =  p]      .     .     .     (2) 
So  these  are  the  conditions  for  contact  of  the  first  order. 

If  m  addition  /"(/>)  =  F"(/>), 

they  are  said  to  have  contact  of  the  second  order,  and  so  on. 


Elementary  Calculus. 


297 


In  general,  a  straight  line  has  only  contact  of  the  first 
order  with  a  curve,  because  the  two  equations  above  (i) 
and  (2)  (one  function  representing  the  straight  line,  the 
other  the  curve),  are  just  sufficient  to  determine  the  two 
arbitrary  constants  for  the  equation  of  a  straight  line,  since 
two  simultaneous  equations  furnish  only  enough  conditions 
to  determine  two  unknowns. 

Likewise  a  circle  requiring  three  conditions  may  have 
contact  of  the  second  order,  for  three  equations  will  then 
be  required,  namely: 


Total  Curvature. 

ART.  72.  The  total  curvature  of  a  continuous  arc,  of 
which  the  bending  is  in  the  same  direction,  is  measured  by 
the  angle  that  the  tangent  swings  through,  as  the  point  of 


Fig.  ay. 


tangency  moves  from  one  end  of  the  arc  to  the  other;  or 
what  is  the  same  thing  it  is  the  difference  between  the  slopes 
at  these  two  points.  In  Fig.  27  the  total  curvature  of  the 
arc  MN  is  (f>f  —  (j)  =  A<£,  say.  It  is  evident  from  geometry 


298  Elementary  Calculus. 

that  </>'  -  0  =  AED.  That  is,  the  total  curvature  is  the 
angle  between  the  two  tangents,  measured  from  the  first  to 
the  second,  hence  it  may  be  either  positive  or  negative, 
according  to  our  conventional  rule  for  positive  and  negative 
angle. 

The  average  curvature  is  the  ratio  between   the   total 

curvature  and  the  length  of  the  arc,  say  -—  &-  ,  where  As  = 
the  arc  length. 

Measure  of  Curvature. 

ART  73.  Following  the  principle  of  minute  increments, 
the  value  of  the  average  curvature,  as  the  arc  becomes 
indefinitely  small,  is  taken  as  the  measure  of  curvature, 
usually  designated  as  *.  But  as  As  becomes  indefinitely 
small,  A^>  likewise  becomes  indefinitely  small,  and  even- 

tually —^  becomes  -^  in  our  notation;  that  is, 
As  as 

K=d$_ 

ds  ' 

Since  tan  6  =  -%-, 

dx 


But  K_ 

uc  ds 


dx 


Elementary  Calculus.  299 

RADIUS    OF    CURVATURE. 

ART.  74.  The  circle  tangent  to  a  curve  (or  having  con- 
tact of  the  second  order)  at  a  given  point  and  having  the 
same  curvature  as  the  curve  at  that  point  is  called  the 
circle  of  curvature  for  the  curve  at  that  point.  In  a  circular 
arc,  the  angle  made  with  each  other  by  the  tangents  at  the 
extremity  of  the  arc  is  the  same  as  the  angle  between  the 
radii  to  these  extremities,  since  a  radius  is  JL  to  a  tangent  at 
the  point  of  tangency,  and  a  central  angle  equals  (in  radians) 
arc  divided  by  the  radius.  But  the  angle  between  the 
tangents  is  the  total  curvature,  A0. 

.'.  A<£  =  = (calling  r  the  radius), 

radius         r 

dividing  by  As, 


As       r 
And  since  r  is  a  constant, 

+ 


ds  r  K  o?y 

dx2 

Since  a  circle  can  always  be  found  of  such  radius  that  it 
will  have  the  exact  curvature  of  any  curve  at  a  given  point, 
the  r  as  found  above  is  called  the  radius  of  curvature  of  a 

given  curve  at  any  point  for  which  —    and  —  «L  are  deter- 

dx  dx2 

mined. 

The  radius  of  curvature  is  understood  to  be  positive  or 
negative  according  as  the  direction  of  bending  is  positive 

or  negative;  that  is,  according  as  —  -is  positive  or  negative. 

dx 


300  Elementary  Calculus. 

EVOLUTE   AND    INVOLUTE. 

ART.  75.  As  every  point  on  a  curve  in  general  has  a 
different  centre  of  curvature,  that  is,  the  centre  of  its  curva- 
ture circle  is  different,  these  centres  describe  a  locus  as 
the  point  on  which  the  curve  moves  along.  This  locus  is 
called  the  evolute  of  the  curve.  It  will  be  seen  later  on  that 
this  name  is  peculiarly  appropriate. 

The  curve  itself  is  called  the  involute  of  its  evolute. 

Involute  arcs  are  used  extensively  in  modern  gears, 
where  the  evolute  is  usually  a  circle. 

ART.  76.     To  find  the  equation  of  the  evolute,  let  the 
curve   equation   be   y  =  f(x)       ........     (i) 

The  equation  to  a  circle  is, 

(x  -  h)2  +  (y  -  k)2  =  r2     .     .     .     .     (2) 

If  this  be  the  curvature  circle  at  the  point  (x,  y)  on 
y=  f(x)t  then  the  x  and  y  in  (2)  have  the  same  value  as 
in  (i)  for  that  point,  by  definition  of  circle  of  curvature. 
Taking  derivative  of  (2  )  twice  with  respect  to  x, 

(*  -  h)  +  (y  -  t)  g  =  o     .    .     .    .     (3) 

<+(£)'  +  <>-*>-&=  °  •  •  «> 

Eliminating  y  between  (3)  and  (4), 

&V1  dyV 


dy  <Py 

dx2  dx* 


dx2  dx2 


As  we  know  r  = 


Elementary  Calculus.  301 

r  +  (Jy 

'  ^ 


If  no  particular  point  on  the  curve  be  taken  fe),   (52) 
and  y  =  j(x)  will,  by  combination,  give  the  equation  of 


the  e  volute  of    y  =  f(x),  -?-  and    -—^-  being  found  from 

ax  ax 

y  =  /(*)- 

Example  :  Find  the  evolute  of  the  hyperbola  xy  =  c2. 

Here          y  =  -     .     .     .     (i)    \y  =  /(*)! 

x 

dy  <? 

whence        -f-  =  —  —j  , 

dx  x 

d*y       2  c2 

and  —  ^  =  —  r  • 

<fo2        ^c3 

Substituting  in  (5^  and  fe), 


From  (2),     h=   -    ±f-+x=  -    ^JLE.    ...     (4) 
From  (3),      k  =   -  -^ h  y  (or  since y=  —  from  (i)) 

.4 

•  •   (s) 


302  Elementary  Calculus. 

Adding  and  subtracting  successively  (4)  and  (5), 
,    ,    ,  _  c6  +  3  g4*2  +  3  c2x*  +  *6  _    (c2  + 

fl  ~T  K  —      -  —  -  —    - 

2  C2X? 


Extracting  cube  root  and  then  squaring, 


Subtract; 


The  equation  to  the  evolute  is  then, 

(h+k)*-  (h-kf  =  (40§, 

where  h  and  k  are  the  general  co-ordinates,  like  x  and  ^ 
in  the  usual  form. 

PROPERTIES  OF  THE  EVOLUTE. 

ART.  77.  An  important  relation  between  evolute  and 
involute  is  the  following:  The  difference  between  any  two 
radii  of  curvature  equals  the  length  o)  the  arc  of  the  evolute 
between  the  two  centres  of  curvature  from  which  they  are 
drawn.  This  important  fact  is  proved  thus: 

Let  (off,  /)  be  any  point  on  the  curve  y  =  /(#);  R,  the 
radius  of  curvature  for  this  point;  (h,  k},  the  correspond- 
ing centre  of  curvature,  and  a  the  angle  R  makes  with  the 


Elementary  Calculus.  303 

#-axis.     Then  the  equation  of  R,  passing  through  (#',  /) 
and  making  angle  a  with  the  *-axis,  is 

y  —  y'  =  tan  a  (x  —  yf)     .....     (i) 

But  R  also  passes  through  (h,  k),  hence  (&,  k)  must  sat- 
isfy (i). 

.-.   (k  -  /)  =  tan  a  (h-  x'), 

k-y' 

whence  —  *—,  =  tan  a. 

h  —x' 


Squaring  and  adding  i  to  both  sides, 
(h  _  *)*  +  0  -  /)2 


(h  -x')2 

But  since  R  extends  from  (h,  k)  to  (xf,  /)  its  length  is 
given  by  Analytics  as,    . 

(h.-  x')2  +  (£-/)2==  R2- 

Substituting  in  (2),  inverting  both  sides  and  extracting 
square  root, 


whence        h  —  x'  =  R  cos  a,  or  h  =  x'  +  R  cos  a  )      ,  , 
and  k  —  y'  =  R  sin  a,  or  &  =  y'  +  R  sin  a  ) 

Differentiating  (3),  [V,  ^r,  R  and  a  are  all  functions  of  yf\. 
dh  =  dxf  +  cos  a  dR.  —  R  sin  a  da  )  ,  _,^ 

dk  =  dy'  +  sin  a  dR  +  R  cos  a  da  ) 

By  Art.  67,  —  =  cos  <p  or  dx  =  cos  0  d 

ds  i  /  v 

and  —  =  sin  0  or  dy  =  sin  (f>  ds 

ds 

Since  the  tangent  to  y  =  f(x)  is  also  tangent  to  the  cur- 
vature circle  at  (x',  /),  R  is  _L  to  this  tangent,  hence 
a  =  90°  +  0,  whence  cos  <j>  =  sin  a  and  sin  <j>  =  —  cos  a. 


304  Elementary  Calculus. 

Also  da  =  d(£>. 

dx'  =  sin  ads. 
Substituting  in  (4),      dy'  =  —  cos  a  ds. 

ByArt.74 
or  since  d(j>  =  da, 


da 
and  (4)  finally  becomes, 


—  =  -  ;  that  is,  ds  =  Rda. 
ds       R 


doc?  —  R  sin  a  da, 

dyr  =  —  R  cos  a  da. 

Substituting  these  values  in  (3d), 

dh  =  RTsirra^  4-  cos  a  dR  —  Rlfa^da  =  cos  a  dR. 
dk  =  —  Ra35-«^/a:  +  sin  a  dR.  +  Rcd&.a  da  =  sin  a  dR 
Squaring  and  adding, 

~dh2+~dk*=  (cos2a+  sin2a)dR2=  dR2 

[since  cos  2  a  +  sin  2  a  =  i]. 

But  (h,  k)  being  a  point  on  the  evolute,  letting  s  be  the 
length  of  an  arc  from  this  point, 


dk2.     (By  Art.  67.) 


or  J5  =  ±  dR, 

which  means  that  R  either  increases  or  decreases,  but  in 
either  case  changes  just  as  fast  as  s. 

It  follows  from  this,  that  the  end  of  a  stretched  string 
unwinding  from  the  evolute  will  describe  its  involute,  or  a 
straight  line  rolling  on  the  evolute  as  a  tangent,  any  point 
on  it  describes  an  involute.  This  latter  method  is  used 
by  draftsmen  to  draw  gear  teeth. 


Elementary  Calculus.  305 

ENVELOPES. 

ART.  78.  The  equations  of  curves,  in  general,  contain 
one  or  more  constants,  and  when  these  constants  vary  the 
result  is  a  family  of  curves,  having  the  same  generic  quali- 
ties, but  differing  in  the  constant.  For  example,  in  the 
equation  to  a  straight  line, 

y  =  mx  +  b. 

If  m  varies,  the  result  is  a  set  of  straight  lines  passing 
through  the  same  point,  (o,  &),  and  making  different  angles 
with  the  x-axis.  Again  in  the  ellipse  equation, 

*+£-!, 

a2         b2 

if  a  and  b  both  vary,  but  always  obeying  the  condition, 
a2  —  b2  =  c2  [c2  being  a  constant], 

the  result  is  a  family  of  ellipses  with  the  same  foci  but 
different  axes. 

The  locus  of  the  intersections  of  consecutive  curves  of  a 
family,  as  the  points  of  intersection  approach  coincidence, 
that  is,  when  the  constant  (or  constants)  changes  by  infini- 
tesmial  increments,  is  called  the  envelope  of  this  family. 

TO  FIND  THE  EQUATION  OF  AN  ENVELOPE. 

ART.  79.  Let  /  (x,  y,  m)  =  o,  be  the  equation  of  a 
curve,  m  being  originally  a  constant.  Then 

/  (x,  y,  m  +  Aw)  =  o 
will    represent    the    curve    immediately    adjacent    to 

/  (x,  y,  m)  =  o, 
Aw  being  indefinitely  small,  when  m  is  allowed  to  vary. 


306  Elementary  Calculus. 

From  I  (x,  y,  m)  =  o          ......     (i) 

and  /  (x,  y,m  +  Aw)  =  o      ....     (2) 

we  get  by  subtracting  and  dividing  by  Aw, 

/  (x,  y,  m  +  Aw)  -  /  (x,  y,  w)  (. 

Aw 

But  by  Art.  62  (3)  may  be  represented  by 


<*'?•  "I  as  AM  A  o. 


hence 


or  more  simply, 

|^=o        ......       (4) 

3w 

By  definition  of  envelope  (4)  represents  a  point  on  the 
envelope,  since  it  is  the  intersection  of  two  consecutive 
curves  f(x,  y,  m)  =  o  and  }(x,  y,  m  +  Aw)  =  o,  as  they 
approach  coincidence,  for  in  (3)  these  equations  were 
combined.  If  now  m  be  eliminated  between  (4)  and  (i), 
we  get  an  equation  free  from  the  variable  w,  but  deter- 
mined by  the  condition  (4),  which  gives  a  point  in  the 
envelope,  hence  the  result  is  the  equation  for  this  envelope. 

The  varying  constant  is  called  the  variable  parameter. 

Example  :  Find  the  envelope  of  the  straight  line  system 
y  =  mx  +  b  where  b  is  determined  by  the  relation 

b  =   —    (p  being  a  constant). 

Hence     y  =  mx  +  -£-',  y  —  mx  —  ^-  =  o; 
m  m 


whence  —_ =  -  *  +  L  =  o, 

om  om  m* 


Elementary  Calculus.  307 

combining,  y=  mx  +  -*— (i) 

m 

and  —  x  +  -%—  =  o (2) 

m2 

To  eliminate  w,  we  get  from  (2), 

W2=  ^  (3) 


squaring     (i),  V  =  w2*2  +  2  £#  +  -£-    .     .     .     (4) 

m 

substituting  value  of  m2  from  (3)  and  (4), 

y2  =  px  -\-  2  px  +  px  =  4  px, 

which  shows  that  the  envelope  is  a  parabola. 

ART.  80.  It  follows  readily  from  the  fact  that  the 
evolute  of  a  curve  is  the  locus  of  its  centres  of  curvature, 
and  that  the  radii  are  all  normals  to  the  curve  (being  J_ 
to  the  tangents  of  each  point),  that  the  envelope  of  the  nor- 
mals to  any  curve  is  its  evolute,  since  these  normals  (the 
radii)  always  pass  through  the  centres  of  curvature,  which 
all  lie  on  the  evolute. 

EXERCISE    XIII. 

i.   Find  the  points  of  inflection  of  the  curve 


x2+  16 

2.  Find  the  equation  of  the  line  through  the  points  of 
inflection  of  the  curve  y  (x2  +  4)  =  x. 

3.  Find  the  radius  of  curvature  of  the  parabola  x2  =  8  y 
at  the  origin. 

4.  Find  the  radius  of  curvature  of 

y2  =  -          -  at  x  =  a. 
2  a  —  x 


308  Elementary  Calculus. 

5.  Find   the    radius    of    curvature    of    the    hyperbola 
4  x2  —  i6y2  =  64. 

6.  Find    the    radius  of    curvature  of    the  hypocycloid 

X%  +  yl  =  al. 

7.  Find  the  evolute  of  the  parabola  y2  =  2  px. 

8.  Find  the  evolute  of  the  hyperbola  xy  =  c2. 

9.  Find  the  co-ordinates  of  the  centre  of  curvature  of 
4*2  +  93,2  =  36  at  (VS,  j). 

10.  Find  the  co-ordinates  of  the  centre  of  curvature  of 
/=  9*at  (3,3). 

11.  Find  the  points  on  the  ellipse  a2y2  +  b2x2  =  a262, 
where  the  curvature  is  a  maximum  and  a  minimum  respec- 
tively. 

12.  Find  the  radius  of  curvature  of  the  cycloid, 

x  =  r  vers-1  -  —  \/2  ry  —  y2  at  the  point  whose  ordinate 

is  2  r. 

13.  Find  the  evolute  of  the  circle,  x2  +  y2  =  r2. 

14.  Find    the  envelope    of   x    cos  3<^>  -f  y  sin  3^  = 
a  (cos  2$)*,  <f)  being  the  variable  parameter. 

15.  Find  the  envelope  of  a  straight  line  in  the  first  quad- 
rant which  terminates  in  the  co-ordinate  axes,  and  makes  a 
constant  area  with  the  axes. 

1 6.  Find  the  envelope  of  a  variable  ellipse  with  constant 
area,  TT  ab. 

17.  Find  the  envelope  of  y2  =  m(x  —  m)  where  m  is 
the  variable  parameter. 


CHAPTER  XI. 
INTEGRATION    AS   A    SUMMATION. 

ART.  81.  Integration  has  been  considered,  heretofore, 
merely  as  the  reverse  of  differentiation.  We  will  now 
consider  its  real  and  much  more  important  meaning. 

Let  <f>  (x)  be  such  a  function  of  x  that  its  first  deriva- 
tive will  be  a  given  function,  /(#);  that  is,  denoting  the 
first  derivative  by  an  accent, 


tt   \        it/  \        <f>(x  —  A 

f(x)  =  <j>'(x)  =  ^       —  -  -^  --  yv  ;    as  A#  =o, 

whence          </>(x  +  Ax)  -  <j>(x)  =  j(x)  A#      .     .     .     (m) 
In  the  language  of  integrals  we  may  write, 

(x)  dx  =  <f>(x).       . 


Suppose  in  <f>(x\  x  to  start  with  a  value  h  and  change  to 
a  value  k,  <j>(x)  would  change  from  (j)(h)  to  (f>(k),  the 
difference  would  be  expressed  by, 


Suppose  again  that  instead  of  one  jump  from  h  to  k, 
x  changes  by  minute  increments,  say  making  n  successive 
changes  of  &x  each,  then  the  successive  steps  would  be, 
<j>(h  +  A*)  -  ^  (h)  =  j  (h)kx  [by  (m)] 

(j>(h  +  2  A*)  -  <£(/*  +  A*)  =  f(h  4- 
<f>(h  +  3  A#)  -  ^(A  +  2  A#)  =  /(/t  +  2 


309 


3io  Elementary  Calculus. 

adding 

<f>(h  +  nhx)  -  <t>(h)  =  /(&)A#  +  f(h 

f(h  +  2  A#)A*  + 
or  since   h  +  nkx  =  k,  by  our   hypothesis  <f>(k)  —  <f>(h) 


The  left  hand  side  of  this  equation  may  evidently  be 
gotten  by  integrating  j(x}dx,  and  then  taking  the  difference 
between  the  values  of  this  integral  when  x  =  k  and  when 

x  =  h,  for  by  hypothesis    I  f(x)dx  =  <j>(x). 
This  is  usually  written 


f 

Jh 


'h 

and  is  known  as  a  definite  integral  as  was  shown  in  a  spe- 
cific case  under  Art.  43. 

The  right  hand  member  is  plainly  a  sum  of  n  terms,  as 
A#  =  o  and  hence  as  n  =  oo  ,  for  there  cannot  be  an  infi- 
nitely small  increment  unless  there  is  an  infinite  number  of 
terms. 

For  brevity  such  a  sum  may  be  indicated  thus: 

V   f(x)  A#  ( V  being  the  symbol  for  summation  j  . 

^h  \  I 

When  Ax  =  o,  this  is  modified  to 
f(x)dx, 


which  brings  us  back  to  our  integral  symbol,  for  we  have 
found  that  this  sum  is  actually  equal  to  the  definite  integral 
of  j(x)dx  (namely,  </>(k)  —  <£(&)),  hence  definite  integra- 
tion is  a  summation. 

ART.  82.     Let  us  see  what  is  the  further  significance  of 
this  series  whose  sum  we  have  been  finding. 


Elementary  Calculus. 


Let  uv  (Fig.  28)  be  any  curve  whose  equation  is  y  =  f(x). 
Divide  the  #-axis  from  the  point  A  to  P  into  n  equal  parts, 


A     D    G     L     P     Q 
Fig.  28. 

calling  OA,  &,  and  OP,  k,  and  the  equal  distances  AD, 
DG,  etc.,  each  A*. 

Then  AB  =  f(h) 

DE  =  f(h  +  A*) 
GH  =/(/*  + 2  A*) 


RP  =/(&  +  «  A*). 

Form  rectangles  by  drawing  parallels  to  the  #-axis  from 
B,  E,  H,  etc. 

The  sum  of  these  rectangles  will  be  less  than  the  area, 
ABRP,  but  can  be  made  to  approach  it  as  nearly  as  we 
please  by  taking  A#  indefinitely  small,  and  hence  n  indefi- 
nitely large. 
The  area  of  BCDA  =  f(h)  A* 

"    "  EFGD  =  )(h  +  A*)  A* 

"     '"  HKLG  =  f(h  +  2 


"     "  RTQP  =  f(h  + 

Adding;  Sum  of  the  rectangles  =  f(h)  A^  +  f(h  + 
+  f(h  +  2  A^)  Ax  +  f(k)  Ax  [since  h  +  n  Ax  =  k]. 

As  Ax  =  o  this   sum   approaches   ABRP,   hence  finally, 
ABRP  =  /(//-)  dx  +  f(h  +  dx)  dx  +  +  + .  .  f(k)  dx.    But 


312  Elementary  Calculus. 

the  right  Hand  side  is  the  same  as  obtained  in  the  last  article 

Ck 

and  shown  equal  to  /    }(x)dx,  hence  , 

Jh 
areaABRP  =     f  */(#)<&. 

The  area  would  be  given  as  well  by  solving  the  equation 
for  x,  say  x  =  F  (y)  and  integrating  /  F(y)dy,  since  the 

rectangles  could  as  easily  be  formed  with  respect  to  the 
^-axis  and  summed. 

That  is,  the  definite  integral  0}  f(x)dx  between  fixed 
limits,  where  y=  f(x)  is  the  equation  of  the  curve,  is  the 
area  bounded  by  the  curve,  the  x-axis,  and  the  two  ordinates 
corresponding  respectively  to  these  limits,  which  are  the 
abscissas  in  this  case. 

Example :  Find  the  area  of  the  parabola  y2  =  8  x, 
between  the  origin  and  the  point  (2,  4).  Here  the  limits 
are  o  and  4,  the  two  bounding  ordinates,  and  we  have, 

rVSxdx  =  \fs  C2x*dx=  §  Vs  ["(2)a  -c/U-. 

Corollary :  Clearly  if  we  reverse  the  limits  we  get  the 
same  absolute  result,  but  with  contrary  sign,  that  is, 


/>*  ph 

\    }(x)dx=  -   /    f(x)dx. 

Jh  Jk 


It  is  also  evident  that  we  can  take  the  area  from  y  =  h 
to  y  =  j  (being  between  h  and  k)  and  then  the  area  from 
y  =  j  to  y=  k,  and  if  the  curve  be  continuous,  the  sum 
of  these  results  will  be  the  same  as  if  we  went  directly  from 
h  to  k.  That  is, 

C*f(x)dx=    P/(*)dfc+    f  */(*)<**• 

J  h  Jh  J) 


Elementary  Calculus.  313 

Thus  a  definite  integral  may  be  readily  expressed  as  the 
sum  of  any  number  of  definite  integrals,  if  the  difference 
between  their  limits  taken  together  equals  the  difference 
between  the  original  limits. 

It  must  be  carefully  observed  that  j'(x}dx  does  not 
become  infinite  between  the  limits.  When  that  occurs  the 
integral  must  be  broken  up  into  parts  leading  up  to  the  gap 
on  either  side. 

ART.  83.  Remembering  that  definite  integration  is  a 
summation  between  the  limits,  if  the  expression  for  the 
length  of  an  arc 


which  represents  any  infinitesimal  arc  whatever  of  the 
curve,  y  =  f(x),  be  integrated  between  the  limits  repre- 
senting the  co-ordinates  of  its  extremities,  the  result  will 
be  the  sum  of  all  the  infinitesimal  arcs  making  up  the 
total  arc  and  hence  the  length  of  this  arc,  that  is, 


/v 


s  being  the  arc  from  abscissa  h  to  abscissa  k. 
Example  :  Find  the  circumference  of  the  circle, 


Taking  derivative;    ~  =  - 


V  r*-x2 

r +r  i         x2   \  *  r 

whence  s  =  2  I        f  i  +  — J  dx  =  2  r  I 


314  Elementary  Calculus. 

It  is  to  be  observed  that  the  limits  —  r  and  r,  which  are 
the  extreme  values  of  x,  give  the  length  of  the  semi-circum- 
ference only,  and  hence  the  factor  2  above. 

SURFACE    OF    REVOLUTION. 

ART.  84.  It  has  been  shown  (Art  69)  that  the  surface 
of  revolution  for  a  variable  point,  (x,  y}  on  an  arc,  is  given 
by  the  formula, 


where  the  revolving  arc  is  indefinitely  small. 

By  the  same  reasoning  as  before,  the  surface  generated 
by  an  arc  of  any  length  will  be  then, 


where  h  and  k  represent  the  abscissas  respectively,  of  the 
two  ends  of  the  arc. 

SOLID    OF    REVOLUTION. 

ART.  85.     In  exactly  the  same  way,  using  the  expres- 
sion found  in  Art.  68  for  solid  of  revolution, 

dv  =  ny2  dx, 
which  represents  an  infinitely  thin  strip, 


v=  Tt    I    y2  dx, 


gives  us  the  volume  between  the  limits  h  and  k. 

ART.  86.  Clearly  we  are  at  liberty  to  divide  a  given 
area  into  strips  as  we  please  and  to  apply  the  same  reason- 
ing to  their  summation,  so  that  any  one  of  the  above  for- 


Elementary  Calculus.  315 

mulae  may  be  expressed   in  terms  of  y,  if  the  limits   be 
determined  according  to  y.     For  example,  we  may  write, 


for  the  length  of  the  arc,  if  a  and  b  are  ^-limits,  etc. 

EXERCISE    XIV. 

i.   Find  the  length  of  an  arc  of  the  cissoid  y2  = 


2  a  —  x 
from  x  —  o  to  x  =  a. 

2.  Find  the  total  length  of  the  cycloid 

x  =  r  vers 

3.  Find  the  length  of  the  hypocycloid  x*  +  y*  =  r*. 

/  -         --\ 

4.  Find  the  length  of  the  catenary  y  =  -  (ea  +  e    a\ 

from  the  origin  to  the  point  whose  abscissa  is  b. 

5.  Find  the  length  of  ay2  =x?  from  (o,  o)  to  (3  a,  ^V^a). 

6.  Find   the   circumference   of   the   circle, 

(*•-  2)2+    (y+  i)'  =  16. 

7.  Find  the  length  of  y  =  log  x  from  x  =  i  to  x  =  4. 

8.  Find  the  area  of  the  ellipse. 

9.  Find  the  area  of  the  circle  in  Ex.  6. 

10.  Find  the  area  of  the   parabola  y2  =  8  x,   between 
the  origin  and  the  double  ordinate  corresponding  to  x  =  2. 

11.  Find  the  area  of  the  hypocycloid. 

12.  Find  the  area  of  the  circle  x2  +  y2  +  2  rx  =  o. 

8  as 

13.  Find  the  area  bounded  by  y2  =  — —  — -,  the  ordi- 
nate a,  and  the  axes. 


316  Elementary  Calculus. 

14.  Find  the  area  bounded  by  the  axes  and  the  line 

2  +  Z--I. 

a       b 

15.  Find  the  area  between  the  X-axis  and  one  loop  of 
the  sine  curve  y  =  sin  x. 

Find  the  surface  generated  by  revolving  about  the  X-axis 
the  following  curves: 

1 6.  The  parabola  y*  =  2  px  from  x  =  o  to  x  =  p. 

17.  The    circle    (x  —  a)2  +  (y  —  4)2  =  25    above    the 
tf-axis. 

18.  The  ellipse  9  x2  +  16  /  =  144. 

19.  The  line  -  +  -  =  i  between  the  axes. 

a       b 

20.  The  catenary  from  x  —  o  to  x  =  a. 

21.  Find  the  surfaces  generated  by  revolving  about  the 
y-axis  in  Examples  16,  18,  and  20. 

Find  the  volumes  generated  by  revolving  the  following 
curves  about  the  X-axis : 


22.  The  ellipse^-  +  -£--!. 

a          b 

23.  The  circle  x2  -f-  y2  =  r2. 

24.  The  hypocycloid. 

25.  The  witch  y=       8  a* 


x2  +  4  a2 

26.  The  line  — h  -  =  i  between  the  axes. 

a       b 

27.  Find  the  volume  generated  about  the  Y-axis  by  the 
ellipse. 

MISCELLANEOUS   APPLICATION. 

ART.  87.     Since  our  determination  of  volume  depends 
on  our  ability  to  divide  our  solid  into  sections,  whose  areas 


Elementary  Calculus.  317 

can  be  generally  expressed,  and  then  summed,  any  solid 
for  which  this  is  possible  may  be  estimated. 

For  example,  let  it  be  required  to  find  the  volume  de- 
scribed by  a  rectangle  moving  from  a  fixed  point,  its  plane 
remaining  parallel  to  its  first  position,  one  side  varying  as 
its  distance  from  this  point,  the  other  side,  as  the  square  of 
this  distance,  the  rectangle  becoming  a  square  5'  on  the 
side,  at  a  distance  of  4'  from  the  point. 

Take  the  line  -L  to  the  plane  of  the  rectangle  through  its 
middle  as  the  X-axis.  Let  v  be  one  side  and  w  the  other, 
then  by  conditions,  x  being  its  distance  from  the  point 
taken  as  origin  at  any  time, 

v  :  x  :  :  5  :  4,  whence  v  =  *£, 

-  X2 

w  :  x2  :  :  5  :  16,  whence  w  =  ^ —  . 

16 

Hence  the  area  of  the  rectangle  at  the  distance  x  (being  any 
point  between  o  and  4)  is, 

25^ 

VW  =  — >* • 

64 

This  area  representing  any  section  of  the  solid,  if  mul- 
tiplied by  dx,  thus  forming  an  infinitesimal  slice,  and 
summed  between  o  and  4,  will  evidently  give  the  total 

volume;  hence  volume  =f|    I    Xs  dx=  ffcty*}*  =  25  cubic 

JQ  ° 

feet. 

Again :  To  find  the  part  of  the  contents  of  a  cylindrical 
bucket  of  oil  remaining  in  it,  after  the  oil  has  been  poured 
out,  until  half  the  bottom  is  exposed  (see  Figure  29). 

Let  EGH  be  any  section  of  the  remaining  contents, 
taken  parallel  to  the  axes.  Take  the  origin  at  the  centre 
of  the  base  and  the  co-ordinate  axes  as  the  axis  of  the 
cylinder  and  a  diameter  of  the  base. 


318  Elementary  Calculus. 

Then  since    EGH  and  DOC  are  similar, 


GH  =  VBG  X  GA  =  vV  -  x2  [where  OG  is  #], 
and  EH  :  CD  :  :  GH  :  OC, 


or 


EH  = 


[where  h  =  altitude  and  r  =  radius  of  base]. 


H 


Fig.  29. 


Hence  area  EGH  =  J  EH  X  GH  =  * 


2      - 


,,        2hr2 

dx=  --  =  contents  remaining. 


EXERCISE   XV. 
MISCELLANEOUS   PROBLEMS. 

1.  Find  the  volume  generated  by  an  isosceles  triangle  of 
altitude,  h,  moving  with  its  plane  always  perpendicular  to 
the  plane  of  a  circle  of  radius,  r,  and  having  always  the 
ordinates  of  the  circle  for  bases. 

2.  What  is  the  volume  generated  when   the   circle  in 
Ex.  3,  is  replaced  by  an  ellipse  whose  axes  are  2  a,  and  2  b? 

3.  Through  the  diameter  of  the  upper  base  of  a  right 


Elementary  Calculus.  319 

cylinder,  whose  altitude  is  h  and  radius,  r,  two  planes  are 
passed,  touching  the  base  at  the  two  extremities  of  a  diam- 
eter. Find  the  portion  of  the  cylinder  between  the  planes. 

4.  Two  right  cylinders  each  of  radius  3  in.,  intersect  each 
other  at  right  angles,  their  axes  intersecting.     Find  com- 
mon volume. 

5.  Find  the  volume  of  a  pyramid  whose  altitude  is  h 
and  area  of  base  B. 

6.  Find  volume  of  a  cone  whose  height  is  h  and  radius  r. 

7.  In  cutting  a  notch  in  a  log,  the  sloping  face  of  the 
notch  makes  an  angle  of  45°  with  the  horizontal  face.     The 
log  is  3  ft.  in  diameter;  how  much  wood  is  cut  out? 

8.  A  right  circular  cone  has  a  small  circle  of  a  sphere  of 
radius  6  in.  as  base,  and  its  vertex  is  at  the  surface.     If  the 
vertex  angle  of  the  cone  is  30°,  what  is  the  volume  of  the 
sphere  outside  the  cone? 

9.  A  square  hole  is  cut  through  the  axis  of  a  grindstone 
for  a  bearing.     The  grindstone  is  18  in.  in  diameter,  2  in. 
thick  at  the  circumference,  and  4  in.  at  the  centre,  and  has 
conical  faces.     If  the  hole  is  3  in.  square,  how  much  material 
is  removed? 


CHAPTER   XII. 
INTEGRATION    BY   PARTS. 

ART.  88.  It  is  frequently  a  great  aid  in  integration  to 
separate  the  parts  of  an  expression  containing  two  factors, 
thus  producing  either  a  re-arrangement  or  a  change  in 
form  of  the  integral. 

This  is  readily  accomplished  by  using  the  formula  for 
differentiating  the  product  of  two  factors, 

d(uv}  =  udv  +  vdu. 
Transposing,         udv  =  d(uv)  —  vdu. 
Taking  the  integral  of  both  sides, 

I  udv  —  uv  —    I  vdu    ....    (B) 

Example  :  I  x2  cos  x  dx  —  what  ? 

Let  x2  =  u  and  cos  x  dx  =  dv 

then  du  =  2  x  dx  and  v  =  sin  x. 

Substituting  in  the  formula  (B), 

I  udv  —      Ix2  cos  x  dx  =  x2  sin  x  —  2     I  x  sin  x  dx. 

Where  the  x2  cos  x  dx  is  now  made  to  depend  upon  the 
integration  of  x  sin  x  dx,  in  which  the  exponent  of  x  is  one 
less  than  in  the  original  expression.  If  we  treat  this  inte- 
gral the  same  way,  using  (B)  again,  letting  x=  u,  du  will 

320 


Elementary  Calculus.  321 

equal  d(x)  =  dx,  which  eliminates  x  from  the  final  inte- 
gral; then 

2  /  x  sin  x  dx  =  —  2  x  cos  x  + 

2  I  cos  #  d#  =  —  2  x  cos  #  +  2  sin  x, 

by  putting  #  =  u  and  sin  x  dx  =  dv, 

whence  dx  =  du,  —  cos  x  =  v. 

.'.     I  x2  cos  x  dx  =  x2  sin  x  —  2     I  x  sin  x  dx  =  x2  sin  x— 

[—  2  x  cos  x  +  2  sin  x]=  x2  sin  #  +  2  x  cosx  —  2  sin  #. 

In  using  the  formula  (B)  no  general  rule  of  application 
can  be  given  for  choosing  the  value  for  u  and  for  dv,  except 
that  they  should  be  so  chosen  that  one  factor  may  be  made 
to  disappear  eventually  or  to  take  such  a  value  that  in 
combination  with  the  other,  it  may  form  an  integrable 
part  of  the  original  expression.  For  example,  in  the 
expression 

I  x2  tan"1:*:  dx, 

dv  can  only  equal  x2dx  since  x2dx  is  the  only  integrable 
part;  tan"1  x  dx  having  no  known  simple  integral,  then 

u  =  tan"1  x,  dv  =  x2dx, 

dx  x3 

du  =    —    —  ,  -v  =  — 

i  zf  x2'  3 

and  I  udv  =    I  x2  tan"1  xdx=  — — — - 


fudv  =  f 


3    , 
=  x  —    — ^-—  [dividing  x3  by  jc2 


322  Elementary  Calculus. 

x  dx 
+  x2 


i    f*  x3  dx        i     C  j          i     f*2 
.'.  —  /  -  =  —    I  xdx  —    • 

3J    i+x2       3  J  6  J  i 


=        -1  log  (i  +  *»): 
6        6 

(*  «  .  7          5C3  tan"1  #   .   jc2       i 

Hence     I  #2  tan  l  x  dx  =  -  -  +  -         - 

J  366 


EXERCISE    XVI. 

Integrate  by  parts: 

I  x  sin  2  x  dx.  9.     /  cot"1  x  dx. 

.     Cexcosxdx.  10.     lxnlogxdx. 

/r 
?x  sin  x  dx.  n.     I  ze  dz. 

r  r  , 

I  x  sec2jc  dx.  I2-     I  y  tarr  y  dy. 

J  J 

/r  log  (x  +  2 ) , 
^  sin  x  dx.  I3-     I  ^  <** 

^       Vx   +    2 

.     Cx  ten-lx  dx.  14-     /    °8 "    "  - 

J  «/    ("  +  i)2 

/2   .-i  j                  r(ioff»)dv 
ar  cot  lxdx.  15.    I  -    ^-j^ . 

8.     /  log  sin  x  esc  #  cot  ^  dx.        16.     lx*coslxdx. 


i. 


6 


INTEGRATION   BY    SUBSTITUTION. 

ART.  89.  An  expression  may  often  be  simplified  by 
substituting  another  variable  for  a  part  of  the  expression 
to  be  integrated.  No  general  rule  can  be  given,  it  being 
largely  a  matter  for  the  exercise  of  originality. 


Elementary  Calculus. 


323 


An  example  or  two  may  aid: 
dx 


Let 

then 

Substituting, 


xVx2  —  a2 

_  I_ 

"  y' 

dx=  - 


^hat? 


_dy_ 


/dx  r  y2  C       dy 

xVx2-a2~    Ji\/i  J  Vi-a< 


y2 


yy 


=  _  L    C        ady    -  =  1  cos-1  (ay) 
a  J  A/I  —  a2  y2        a 


i        _t  a,       i        _i  x 

—  cos  1  —  =  —  sec  l  —  . 


Again;  /  - 

J  3  x2  ~ 


dx 


2  #  + 


a  a 

=  what  ? 


/dx        _=     r         3  dx 
3*2-2*+f       J  gx2-6x+  5 

[multiplying  and  dividing  by  3]. 
Let  (3  x  —  i )  =  y,  then  dy  =  3  dx  and 


The  suggestion  (3  x  —  i )  =  y  comes  from  the  fact  that 
9  x2  —  6  #  +  5  can  be  put  in  the  form, 

9  ^2-6^+i  +  4 


324  Elementary  Calculus. 

and  the  formula 

-  =  —  tan"1 —  is  immediately  suggested. 

a2  +  x2      a  a 

ART.  90.     Expressions  containing  the  form  \/, 
can  usually  be  integrated  by  making  the  substitution, 


\/x2  +  ax  +  b  =  y  —  x. 


Example :  I  — = 

J  Vx 


=  ? 


+  X-   2 


Let  \x2  +x—  2  =  y  —  x. 

x2  +  x  —  2  =  y2  —  2  yx  +  x2'f 

whence  x  = 


_  y2  +  2 


2  y 


doc  =  2  y  +  4  y2  -  2  y2  -  4  a        2  (y2  +  y  -  2  )  , 
(i  +  2;y)2  (i  +  2;y)2 


/— 

V  ^2 


—  2  =  y  —  x  =  y  —  •— 

i  +  2y 

_  y  +  2  y2  —  y2  —  2  _  y2  +  y  —  2 
i  +  2y    '  i  +  2y 


+  x  —  2        J     y  +  y  - 


=  r 


23; 


=  log  (i  +  2  ff  +  2\/X2  +  X  —  2). 


ART.  91.  Expressions  containing  the  form  \/—x2-\-ax+b, 
where  —  x2  +  a^  +  b  can  be  resolved  into  two  first  degree 
factors,  can  be  integrated  by  making  the  substitution, 


V—  x2  +  ax  +  b  =  \/ (m  —  x}  (n  —  x)  =  (m  —  x}y 


Elementary  Calculus.  325 

or  (n  —  x)y,  where  (m  —  x) 

and  (n  —  x)  are  the  factors  of  —  x2  -f  ax  +  6 

Example :  C  xdx =  ? 

J  \/2  +  3  #  —  2  #2 

V2  +  3  x  —  2  #2  =  \/(i  +  2  #)  (2  —  #)  =  (2  —  x)y, 

i                                    2V2  —  i     j           10  y  dy 
whence  x=  — ,  dx=  22' 

EXERCISE    XVII. 

Integrate  by  substitution: 

1.  I  — [substitute  z3  for  x]. 

J  x*  +  i 

2.  I  —^  [substitute  z6  for  ^]. 
J  x*  +  ^ 

3- 


ar  — 
4- 


^     /f~^~i' 
5.      I  — ^    ^  [substitute  \/y2  +  i  =  z]. 

J     \/yl    _j_     j 

6        C__xdx_  [substitute  a2  -  x2  =  z3]. 

*/  (a2  —  ^c2)* 

/^^ 
v  ^2  ~(~  i  —  i 

s.  r — ^^ — r . 

t/     I         ^         2  X 


f  JZ 

9.          I   --  ==r 

J  Z2\/Z2  —  2 

.     C^4y-y2 

tJ  M2 


10 


326  Elementary  Calculus. 

ii.  r      dx 

•^  #V5  #2  +  4X  —  i 

/x  dx 
-7=  a* 

V2  +  5*-  3* 

13  ^ dx  [substitute  x  = 

J  x\/4  —  x2 

14  /     x  dx —  [substitute  x  —  i  =  z]. 

J  (x-  i)4 

j ,-        /  — ±? [substitute  ez  =  x]. 

J  e2*  -  2  *" 

J  j(V^4  +  ^2  +  i     L  ^     J 

/»<£» 
~x?' 
^i  vVy 

/»\/^    +       I        7 
l8'  /       ~~T 

•/  V  x  —  i 

J  V  2  ax  —  x2 
2CX       /  VJIP  —  x2  dx. 

REDUCTION    FORMULA. 

ART.  92.     Integrals  of  the  general  form 

I  xm  (a  +  bxn^p  dx 
are  exceedingly  common,  as 

J  J    (a2  —  x2)*         J  \/2  ax  —  x2 

Take  for  example, 

r      Xs  dx       ^ 

J     (a2  -  x2}* 


Elementary  Calculus.  327 

x?  dx 


can 


r    x*  dx 

A  careful  inspection  will  show  that  if     I  —       — 5 , 

J  (a2  —  x2y 

/x  dx 
— -,  the   expression    is 
(a2  -  x2Y 

integrable,  for  the  latter  integral  is  in  the   form  xndx  or 
can  be  readily  reduced  to  it   by  inserting  the  factor   2. 

/dx 
-  can  be  found  if  it  can  be  made  to 
(a2  -  x2)* 

C      dx  .  _!  x 

depend  upon  I  —    =  sin  l  —  - 

In  the  former  case  the  exponent  of  x  when  the  expres- 
sion is  in  the  form  /  xm  (a  +  bxnYdx  is  to  be  decreased, 

and  in  the  latter  the  exponent  of  the  parenthesis  is  to  be 
decreased. 

If  then  a  general  method  can  be  devised  for  expressing 

/  xm  (a  +  bxnY  dx  in    terms    of    other   integrals    where 

m  or  p  (or  both)  is  increased  or  decreased  as  the  case  may 
require,  many  of  these  forms  can  be  integrated. 

The  process  in  one  case  will  suffice  to  show  how  these 
formulae,  four  in  number,  known  as  reduction  formula, 
are  found.  The  formula  for  integration  by  parts  is  used, 
as  it  is  necessary  to  break  up  the  original  expression. 

In  I  xm  (a+  bxnY  dx,  then, 

let      u  =  xm~n+l  and  dv  =  (a  +  bxnY  xn~l  dx  [xmdx 


Substituting  in   I  udv  =  uv  —  I  vdu (B) 

tm(a  -f  bxnY  dx  = 


/«• 


xm-n+l 


nb  (p  +  i) 


328  Elementary  Calculus. 


n+  I      Cxm-n  (a 
p  +  i)    J 


dx 
nb  (p 

(a  4- 
Since  du  =  (m  —  n  +  i  )  jcm~n  dx  and  z;  = 


nb  (p  +  i) 

But     /  xm~M  (a  +  bxn  Y+1dx  =   I  x  m~n  (a  +  bxn  )  (a  +  fon  /</«; 

[since  zp+1  =  z.zp] 
=  a    ixm-n  (a  +  bxny  dx  +  b   I  xm  (a  +  bxn}p  dx 

[multiplying  out]. 
Substituting  in  (i)  above, 

f*»  (a  +  bxnydx  =  xm'~+1  (a+^r1  - 
J  nb  (p  +  i) 

I)    /V-  (a  +  b 
i)  J 


nb  (p  + 

b(m—n+i) 
nb  (p  + 
Transposing  the  last  term  of  (2 )  and  collecting, 

— -- -*- • I  xm  (a  +  bxnY  dx  =  - 

nb  (p  +  i)     J  nb  (p  +  i) 

/  j  ^        f* 

—         '•^          n  ~\-    I )     I      m_n   /        i     Tj^nV  /7ir 

• I    vV  1C*-      |^    t/.A'      J      W-vV. 

nb  (p+i)    J 

b  (np  +  m  +  i) 
Dividing  by  rttf+x)      ? 


C 


^w(a+^TC/^= 


^  +   W  +    i) 

.     .     (A) 


b(np  +  m+ 

Here  jcm  (a  +  Zww  /  ^  is  plainly  made  to  depend  upon 
the  integral  /  xm~n  (a  +  bxn)  dx,  which  is  exactly  like 
it  except  that  the  exponent  of  x,  [m],  is  reduced  by  n. 


Elementary  Calculus.  329 

The  other  three  formulae  are  as  follows: 

//v*W2-     (rt     A       /i-vW^ 
x  (-a  +  bx' 


np  +  m 


np  +  m  +  i 
-  Cxm  (a  +  bxn)P-1  dx (B) 


C 
J 


a(m+  i) 

_   b  (np  +  n  +  m+i)  Cm+n  ft         ^ 

a  (»»  +  i)         J 

CX™  (a  +  b 
J 


"+*    Cxm  (a 
i          J 


aw  (^  +  i) 

(A)  decreases  m  by  w. 
(B  )  decreases  p  by  unity. 

(C)  increases  m  by  w. 

(D)  increases  p  by  unity. 

In  using  these  formulae,  the  expression  to  be  integrated 
is  carefully  inspected,  and  the  known  integrable  form  to 
which  it  is  to  be  reduced,  is  decided  upon,  then  the  formula 
[(A),  (B),  (C),  or  (D)]  suited  to  this  reduction  is  applied. 
Clearly  these  formulae  may  all  be  applied  to  one  example 
successively,  or  any  one  of  them  may  be  used  any  number 
of  times  until  the  desired  form  is  reached.  These  for- 
mulae fail  when  the  constants  have  such  a  value  that  the 
denominators  of  the  fractions  reduce  to  zero.  For  ex- 
ample, in  (A)  b  (np  +  m  +  i)  must  not  reduce  to  o,  etc. 


Example:  I  o 


330  Elementary  Calculus. 

Here  the  form  desired  is  plainly 


Va2  -  x2  a 

To  accomplish  this,  x2  must  reduce  to  x°  =  i  and  (a2  —  x2)^ 
must  reduce  to  (a2  —  x2)~*.  That  is,  m  must  be  decreased 
by  2  and  p  by  i  (why  can  it  not  be  reduced  to  the  form 

— °°  x        .     To  accomplish  this,  (A)  must  be  used  to 
Va2  -  x2 
reduce  xm  to  xm~n,  and  (B)  to  reduce  p  to  p  —  i. 

Comparing       I  x2  VV  —  x2  dx  =        I   xi  (Q?  —  x2)*  dx 

with    Cxm  (a  +  bxn)pdx 

m  =  2,  n  =  2,  p  =  1,  a  =  a2,  b  =  —  i 
using  (A)  then, 

/x  (a2  -  x2)$ 
x2(a2  —  x2)%  dx=  — > ' — 
-  4 

2         f* 

-  I  (a2-  x2)*  dx (i) 

-  4  J 

[since  xm~n  =  x2~2=  XQ  =  i]. 
Applying   (B)   to  I  (a2  —  x2)*  dx,  where  m  =  o,  n  =  2, 


—    f \a2  - 


«i*-^ 


2  - 


2 

*  (a2  -  *2' 


Elementary  Calculus.  331 

Substituting    this  value    of     /  (a2— x2)*  dx  in  (i), 

pV*  = 


8  a 


where  I  x2  \/a2  —  x2  dx  is  completely  integrated.  The  value 

of  these  formulae  lies  in  the  ability  to  see  the  integrable 
form  that  lies  within  the  original  expression,  and  to  select 
the  appropriate  reduction  formula.  It  is  a  matter  for 
observation  and  ingenuity  purely. 


Again  I  \/2  ax  —  x2  dx  =  what? 

/dx 
—  - 

V2  ##  —  x2 


dx  ,  x 

=  vers-1  - 


To  put  I  \/2  ax  —  x2  dx  in  the  form  I  xm(a  +  bxn)p  dx, 
take  out  x  from  under  the  radical,  and  we  have 

/  xfc  (2  a  —x)^dx. 
This  must  be  reduced  to 


r 

J 


2  ax  -  x  x     2  a  -  x* 

Since  w  =  i,  here  ^m~n  =  jc^-1  =  x~*  the  desired  form 
for  x,  hence  (A)  is  needed.  Also  p  is  to  be  reduced  to 
p  —  i.  [i  —  i  =  —  J]  hence  (B)  is  also  needed.  Apply- 
ing these  successively  we  get  the  desired  form.  Only  prac- 
tice and  experience  can  give  facility  in  the  use  of  these 
formulae,  and  familiarity  with  the  simpler  integral  forms 
is  desirable,  that  the  inspection  of  the  expression  to  be 
integrated  should  be  effective. 


332  Elementary  Calculus. 

EXERCISE   XVIII. 
Integrate: 

1.  C(x2  +  62)i  dx.  7.    A/2  ry  -  y2  dy. 

2.  '  I  vV2  —  x2  dx.  n       xdx 

J  8    l~7= ;  • 

/>  J  V2  ax  —  x2 

»j- *•-•**     ^ 


9- 


5 


r     dx 

J  ^  v!-:: 

r     dz  r     dx 

'  J  (a2-  z2)*  '  ''  J  Va2  - 

'  f  (*-?*„+&  [substitute  first  •-* 

.  r- 

J  \/I   - 

.  ryZ 

J 


X2 


I  <.      I  Vl   —   2  Z  —  Z2  ffz. 

J  i     r    ^^ 

16.    /Vy2  +  6  Jy.  ^       2  ^  -  ^2 


RATIONAL  FRACTIONS. 

ART.  93.     If  the  fractions  —  $  —  and  -  5  —  be  added 

i  -  x          2  +  33; 

together,  we  get, 


i  —  x      2  +  3^       (i  —  x)  (2  +  $x)       2  +  x  — 


Elementary  Calculus.  333 

It  will  be  observed  that  the  numerator  of  the  sum  gives 
no  indication  of  the  numerators  of  the  component  fractions, 
but  that  the  denominator  does  indicate  directly  the  denomi- 
nators of  the  components.  If  the  denominator  is  in  the 
form  indicated  in  the  final  fraction  above,  it  is  easy  to 
factor  it. 

So  that  we  may  regard  every  rational  fraction  whose 
denominator  is  factorable  as  made  up  of  simpler  fractions 
having  respectively  the  factors  as  denominators.  If  it  is 
required  to  integrate,  for  example, 

11  +  **     dx, 


2  +  X  -  3  X2 

it  is  clearly  a  gain  to  be  able  to  express  this  fraction  as  the 
sum  (algebraic  sum  of  course  is  meant)  of  two  or  more 
simpler  fractions;  for  when  we  discover  that, 

ii  +  ^x      =      3        j         5 

2  +  X  —  3  #2          I   —  X         2  +  3  X 

we  get  the  integral  readily,  since 

=  -3  log  .(i-#)  and  C-^-  =  $\Qg  (2  +  3*). 
J  2  -f-  3  x     3 

Since  we  know  that  this  decomposition  is  possible,  for 
every  denominator  factor  we  set  a  fraction  with  a  letter,  or 
letters,  for  numerator,  which  we  determine  by  the  principle 
of  identities. 

It  is  necessary  to  discriminate  between  first  degree  and 
second  degree  factors,  as  will  appear,  hence  we  have  four 
cases,  as  follows: 

(a)  where  the  factors  are  linear  only,  and  not  repeated. 

(b)  where  the  factors  are  linear  and  repeated. 

(c)  where  the  factors  are  quadratic  and  not  repeated. 

(d)  where  the  factors  are  quadratic  and  repeated. 


334  Elementary  Calculus. 


Case    (a). 
r  in  t 
component  fraction  of  the  form 


For  every  linear  factor  in  the  denominator  there  is  a 

A 


x  ±  a 


Suppose  the  fraction  is    /v   ' •  ;  where  F  (x)  =  (x  ±  a) 
(x±b)  (x±c)  .  .  .   (x±n). 
Then 

/(*)  =       ^         ,        B  C  N 

JF(#)       (a;  ±  a)       (x±b)       (x  ±  c) '  x  ±n 

The  original  fraction  should  be  a  proper  fraction,  that  is, 
the  degree  of  the  numerator  should  be  less  than  that  of 
the  denominator,  to  avoid  complications.  If  this  is  not  the 
case  in  the  given  fraction,  it  can  be  made  so,  by  dividing 
numerator  by  denominator  until  the  remainder  fraction 
fulfills  this  condition.  The  remainder  is  then  decom- 
posed and  the  integral  quotient  added  to  the  result.  An 
example  will  make  the  process  plainer: 

(x2-  i)  dx 


X2  —    I  X2  —  I 


(X*  -  4)   (4  *2  -   I)          (X  -  2)  (X  +  2)   (2  X  -  I)  (2  X  +  I) 

A  B  C  D 

~  X  —   2          X+22X—    I2X+I 

It  is  to  be  remembered  that  this  is  an  identity,  not  a  mere 
equation,  as  the  two  sides  must  be  exactly  the  same,  when 
cleared  of  fractions  by  our  hypothesis,  A,  B,  C  and  D  being 
used  because  we  do  not  immediately  know  what  their 
values  are. 


Elementary  Calculus.  335 

Clearing;  x2  -  i  =  A  (x  +  2)  (2  x  -  i)    (2  *  +  i)  +  B 

(X-2)    (2X-I)    (2X+   l)+   C  (X-2)  (X+  2)    (20C  +  i)  + 

D  (#  —  2)(#  +  2)  (2  x  —  i).  Since  this  is  an  identity  it  is 
true  for  any  value  of  x  whatever;  hence  we  can  give  x  such 
values  that  the  terms  will  all  disappear  but  one,  and  thereby 
find  the  unknown  constant  it  contains.  For  example,  if  we 
let  x  =  2,  all  the  terms  containing  (x  —  2)  will  reduce  to  o, 
hence 

22  -  i  =  3  =  A  (4)  (3)  (5)  +  o  +  o  +  o  =  60  A, 

whence  A  =  ^V 

Let  x  =  —  2,  and  all  terms  containing  x  +  2  will  reduce 
to  o  ;  hence  (-  2)2  -  i  =  3  =  o  +  B  (-  4)  (-  5)  (-  3) 
+  0  +  0=  -  60  B, 

whence  B  =  —  ^j. 

Let  x  =  \  ;  then 

(i)2  -i-    -  1=  o  +  o  +  C  (-  f)  (f)  (2)  =   -  V-C, 
whence  C  =  +  TV 

Let  x  =  —  i  ;  then 

-  (i)2-  1  =  -  I  =o  +  o  +  o  +  D  (-  f  )  (|)  (-  2)=  V-  D, 
whence  D  =  —  TV 

dx 


Then     r  _  (x2  -  1)  dx     =  _i_  r  dx     j_  r_ 

J    (X2  —  4)   (4X2  —   i)          20  J   X  -    2         20JX 

+  j_  r  <**   _  JL 

10  J    2  X  —   I  IO 


+2 
=   A"  ^g    (*-«).-  A   log   O  +   2)  +  A  ^g    (2  *  -    I) 

-  A  log  (2*  +  J)« 

j   ,       (jc—  2)  (2  x—  i)  (by  the  principles  of 

""  "  °§  (^  +  2)  (2  *  +  i)  logarithms.) 


336  Elementary  Calculus. 

Case   (b). 

In  using  indeterminate  coefficients  of  any  sort,  it  is  a 
cardinal  principle  that  every  possible  case  that  may  arise 
must  be  provided  for  in  the  supposition  used. 

Suppose —-2 — ,       5  ~  x   ,    and   -^—    ^  are  added, 
i  —  x       (i  —  x)2  (i  —  x)3 

3  5  —  x      _   3  x2  +  i        7  —  12  x  +  x2 

i  —  x       (i  —  x)2       (i  —  x)3  ~        (i  —  x)3 

Here  the  (i  —  x)s  gives  no  indication  directly  of  the 
factor  (i  —  x)2,  that  has  disappeared  in  it.  If  (i  —  x}3  is 
separated  into  linear  factors  they  would  all  be  alike  (i  —  x), 
(i  —  x),  (i  —  x),  and  there  would  be  no  separation  at  all, 
neither  would  the  fractions  having  denominators  (i  —  x)2 
and  (i  —  x)3  be  provided  for.  That  nothing  may  be 
omitted  it  is  necessary  then  to  provide  a  fraction  for  each 
of  these,  hence  for  every  factor  of  the  form  (x  ±  a)n  a 
series  of  fractions  is  assumed,  thus: 

/(*)  A          ,   B 


(x±a)»-  (*±*r 

thus  accounting  for  all  the  powers. 
Example:  Cx5  ~  S^-^=  ? 

As    this    is    an    improper    fraction,    divide    numerator    by 
denominator, 


/r5_53.2_3 

r 

2fdx  +  3 

c 

X2(X+I}2 

x3  —  x: 

dx  —    1  x  dx  — 
!-  i       A    ,   B 

x2  (x  + 

\9                      9       ' 

•  iV       x*      x 

'    (*+i)2 

dx 


+     D 

^T^Ti 


Elementary  Calculus.  337 

[Thus  accounting  for  all  the  powers  of  x.  and  of  (x  +  i).] 
Clearing; 


Let  x  =  —  i  ;  then 
(-  i)3  -  (-  i)2  -  i  =  -  3  =  o  +  o  +  C  (-  i)2  +  o=  C, 

C=-3- 
Let  x  =  o;  then 

o  —  o  —  i=  —  i  =  A(i)2  +  o  +  o  +  o=A 
A-  -  i. 

Since  no  rational  value  of  x  will  cause  the  other  terms  to 
disappear,   we   will   give  x  any  small  values   to   get  two 
simultaneous  equations  for  the   two   remaining  constants, 
B  and  D. 
Let  x  =  i  ;  then 

i3  -  (i)2  -  i  =    -  i  =  A  (2)2  +  B  (i)  (2)2  +  C  (i)2 
+  D(i)2(2), 

or  since  A  =  —  i,  and  C  =  —  3 

-i=    -4  +  4B-3  +  2D 
whence  2B  +  D=3  .......     .     (i) 

Let  x  =  2  ;  whence 

3B+2D=4     .......     (2) 

Combining  (i)  and  (2) 

B  =  2  and  D  =  —  i. 

Hence, 

dx  Cdx  dx 


x3  -  x2  -  i  , 

^^TWdx' 


r_dx_=L+  2  j     +  _JL_  _  iog  (x 

J   X+   I         X  X+    I 


338  Elementary  Calculus. 

[collecting]. 


Case   (c). 

If  for  a  factor  of  the  second  degree  we  set  a  fraction  of 

^ 
the  form  -     —  ,  we  overlook  the  possibility  of   the 

x2  +  a  x  +  b 

form—  —  -  -  ,  since  this  is  also  a  proper  fraction,  but 

x2  +  ax  +  b 

if  both  are  combined  in  one  thus  getting  the  most  general 
form,  all  contingencies  are  provided  for.     So  for  factors  of 
the   form   x2  +  ax  +  b,   we    have   fractions   of   the   form 
Ax  +  B 


x2  +  ax+  b 
Hence     -* 

where         <£ 
Example: 

M.       * 

w  -t  B      |      ( 

:*  +  D 

(x)      x2.- 

(X)  =    (X2 

r 

|-  ax  +  b  '  x2 
+  ax+b)  (x2 

2X2+   I 

+  cx+d 
+  cx  +  d) 

1  (v 

*J   vr* 

A 

L+c 

\(K  4-    T^l    l'c 

(^+l)(^2+l)          X+l  X2+I 

Clearing; 

2x2+i  =  A(^2+i)+  (x+  i)(B*+[C)    .....   (i) 

It  is  plain  that  no  rational  value  of  x  will  make  x2  +  i 
equal  to  zero,  and  in  general  with  quadratic  factors  this 
process  is  useless.  Either  x  can  be  given  any  arbitrary 
values  as  in  the  last  case  or  the  following:  method  be  fol- 


Elementary  Calculus.  339 

lowed;   a  method  that  is  entirely  general  and  can  be  used 
in  every  case  if  preferred. 
Multiplying  out  in  (i); 

2  x2  +  i  =  Ax2  +  A  +  Ex2  +  Cx  +  "Bx  +  C. 
Collecting; 

2  x2  +  i  =  (A  +  B)  x2  +  (C  +  B)  x  +  (A  +  C). 

Since  this  is  an  identity,  the  coefficients  of  like  powers  of  x 
on  the  two  sides  are  identical;  that  is, 

A  +  B  =  2     coefficients  of  x2. 
C  +  B  =  o    since  there  is  no  x  on  the  left. 
A  +  C  =  i     absolute  terms. 
Combining  these  as  simultaneous: 

B=i,  A=|,  C=-  j. 


.     r    (2x2  +  i}dx    _ 

3    f   dx      {  i    Cx-  i 

'  J  (x+  i)(x2+  i) 
_  .,     C   dx          i 

2j    X  +  I           2j#2+I 

r  ^^     i  r  dx 

2  J  X  +   I     '   2* 

1  X2  +   I        2j     X2  +    I 

Case   (d) 

The  same  reasoning  that  was  used  in  case  (b),  will  show 
that  for  every  factor  of  the  form  (x2  +  ax  +  b)n  there  is  a 
series  of  fractions  with  numerators  of  the  form  A#  +  B 
and  denominators  successively,  (x2  +  ax  +  b}n, 
(x2  +  ax  +  b)n~\  (x2  +  ax  +  b)n~2  .  .  .   (x2  +  ax  +  b). 

Example: 

x2  -  2  x  +  $      A_      B      Cx  +  D      &y  +  F 

x       (x2+2)3 


34°  Elementary  Calculus. 

EXERCISE    XIX. 

Separate  into  rational  fractions  and  integrate: 

2  x  -  3       , 
•  a 


*   -   3 

6        - 


x?  —  6x2  + 


21 


C    3  x  -  i      J 
'     J   J+ *-,**• 

22.  r   6*+* 

J    (X+  2)3(X-  I) 

23.  r-^j_</, 

J    Z3  +    2  Z2 


2  <<• 


r 

J 

.  r  (^-6)^  .        ^  J 

J 


*   -  3 

r  2^- 


/«3  „     \ 
vs 

9-  A — s 

»y      (^^V  I J   ^, 

l8.  /  <ty. 

/*  ^2  _  _|_    2 

20.          I    ^-^ +  4  ^    — _I_  ^^ 

J  (^2+3)3 


+  i 


Elementary  Calculus,  341 

r * 

J  (x2+  i 


I      Q  ,  N2         2     i  \ ^A' 


CHAPTER  XIII. 
TRIGONOMETRIC    INTEGRALS. 

ART.  94.  The  integration  of  the  more  complex  trig- 
onometric functions  can  often  be  accomplished  by  substi- 
tution, sometimes  by  breaking  up  the  expression  taking 
advantage  of  the  relations  known  to  exist  between  the 
different  functions.  There  are  very  few  general  rules  and 
the  chief  assets  are  originality  and  a  knowledge  of  the 
simpler  integrable  forms.  A  few  cases  may  be  noted, 
however. 

ART.  95.     Integrals    of    the    form     /    smmxcosnxdx 

where  either  m  or  n  is  a  positive,  odd  integer. 
Say  m  is  odd;  then  since  sin2#  =  i  —  cos2  x, 


m-l 


I    sinm  x  cosn  x  dx  =    /    ( i  —  cos2  x)        cosn  x  sin  x  dx 

m-l 

/~~2 
(i  —  cos2#)       cosn  xd  (cos  x). 

m-l 

A 

[For  sin  m  x  =  sinm"     sin  x  =  (i  —  cos2  x)        sin  x.] 

Since  m  is  odd,  m  —  i  is  even  and  hence  — can  be 

2 

expanded  by  the  binomial  theorem;  then  each  term  mul- 
tiplied by  cosn  x  d  (cos  x)  becomes  an  integral  of  the  form 

342 


Elementary  Calculus.  343 


and  the  result  is 


/x  njx  =  ^L  t  or    f  d-2.  =  log  *, 
»  +   I  J        X  _ 

easily  found. 

If  n  is  odd,  the  cos  x  is  reduced  to  sin  x  and  the  same 
process  followed. 

C  cos3*    ,          -> 

Example:  -  dx  =    ? 

J    sin* 

[«**.&=  f  i^sini*  cosxdx=  fjfisin*) 

J    sin  *  J        sin  *  J      sin  * 

_     /    sin  x  d  (sin  *)  =  log  sin  x  —  4  sin2  x. 

If  w  +  n  is  an  even  negative  whole  number, 

sinw  x  cosn  ^  may  be  put  in  the  form 


I 

cosUoc.  sin  ™+"xdx=    C  cotn  x  csc-<w  +  ^  x  dx, 
sinn  x  J 

smmx  C05m  +  nx(ix==    I 
cosm  x  J 


or 


Since  m  +  w  is  an  even  negative  integer,  —  (m+n  ) 
will  be  a  positive  even  integer,  hence  leaving  sec2#d#  as 
the  d  (tan*),  sec  -  (m+n~>  -2  x  can  be  expressed  entirely  in 
terms  of  the  tangent  by  the  relation  sec2  x  =  i  +  tan2  x. 

Example:     Cc-^-^dx=?     Here  m  +  n  =  -  6  +  2  =  -4. 
J  sin6* 


Hence 


C  ?2£*  fc  =  f  £2^  sin-.  xdx=   C  cot*  *  csc<  *  dx. 
J    sm°  *  «y    sin2  *  J 

The  cot2  x  +  i  =  esc2  x,  hence, 

I    cot2  *  esc4  *  dx  =    I    cot2*  (i  +  cot2*)  csc*xdx 


344  Elementary  Calculus. 

J   c°  ~T~  ~7~ 

ART.  96.     If  the  integral  is  in  the  form,    /  sec2m  *  dx  or 

I    csc2nxdx,  where  n  and  m  are  positive  integers,    the 
expressions  can  be  readily  put  in  the  forms, 


=  (tan2*  +  i)  m~l  d  (tzn  x) 

2  n  —  2 

and  (cot2  *  +  i )      2      esc2  *  dx 

which  are  both  readily  integrable,  since  m  —  i  and  n  —  i 
are  both  integers  and  the  parentheses  may  be  expanded. 

Example:  I  — ^—  =    ? 

J    cos6* 

/dx       r  r 

cos6  x       J  J 

=    i  (tan2  *  +  i  )2  d  (tan  *) 

=   /   tan4  *  d  (tan  *)  +  2    /   tan2  *  d  (tan  *)  +    / 

tan5  *   ,     2  tan3  *    . 
h  —        -  +  tan  *. 

5  3 

ART.  97.     If  the  integral  is  of  the  form,      , 

I   secm  *  tann  *  dx  or    /    csc™  *  cotn  *  dx, 


sec 


Elementary  Calculus.  345 

where  m  is  anything,  and  n  is  a  positive  odd  integer,  it  may 
be  reduced  to 

f* 
secw-i  x  tan™-1  x  sec  x  tan  x  dx 


=    /    sec™-1  xt&n71-1  xd  (sec#), 


or 


I     csc^-^cot™-1^  (csc#), 

and  since  n  is  odd,  w  —  i  is  even  and  tan  x  and  cot  x  can 
be  expressed  in  terms  of  sec  x  and  esc  x  respectively  by 
the  relations,  tan2#  =  sec2#  —  i  and  cot2^  =  csc2#  —  i. 
ART.  98.     If  the  integrals  are  in  the  forms, 

/   tanm  xdx  or  I    cotm  x  dx, 
they  may  be  put  in  the  forms, 

I    tanm~2  x.  tan2  xdx  —    I    tanm~2  x  (sec2  x  —  i  )  dx, 

and       I    cotw~2  x.  cot2  xdx  —    I    cotm~2  x  (esc2  x—  i  )  dx. 

If  these  are  multiplied  out,  the  first  term  is  always  inte- 
grable  and  the  exponent  of  tan#  or  cotx  is  reduced  by  2 
in  the  second  term;  thus  each  application  of  the  process 
reduces  the  exponent  m,  until  an  integrable  form  is  reached. 

Example:  I   (t&n4x)dx=   ? 

tan4  xdx  =    I     tan2  x  (sec2  x  —  i)  dx 


346  Elementary  Calculus. 

=   I   tan2  x  d  (tan  x)  —     I    tan2  x  dx 
—    I    (sec2  x  —  i )  dx 
CsK*xdx+    C  dx 


tan3  x 

3 
tan3 


3 

—  tan  x  +  x. 


3 

ART.  99.     When  m  and  n  are  both  positive  integers  the 
multiple  angle  formulae  may  be  used  to  simplify,  namely, 


Example :          I  sin4  x  cos2  x  dx  =  ? 

I  sin4  x  cos2  x  dx  =    I  (sin  x  cos  #)2  sin2  x  dx 


sin2  2  x  dx  —      I  sin2  2  #  cos  2 


=  J  /  sin2  2  x  dx  —  \  I  si 
~  iV  I  (x  ~~  cos  4x)  dx  —  TV  I  sin2  2  je  cos  2  #  (/  (2  rv 
=  TV  I  dx  —  ^  I  cos  4  jc  J  (4  x)  —  TV  /  sm2  2  xd  (s*n 
=  TV*-  &  sin  4  #  -  &  sin3  2  5f. 


Elementary  Calculus.  347 

ART.  TOO.     The  following  formulae  will    be  useful,  but 
their  derivation  is  not  necessary  here. 


dx  *  «     i 

where  m  >  «, 


The  integration  of  -     —  ;  —    Js  made  to  depend  upon 
m  +  nsmx 

the  same  form  by  first  substituting  x  —  z  +  90°. 
sin  g^Csin^  -  n  cos 


ea*  cos  w^  ^  =  gffjfcsjn  na;  +  g  cos 
a2  +  w2 

EXERCISE  XX. 


.       Icsc4xdx.  9.       It 

Aan3  jc  </*  /' 

-  10.      /  ( 

J      cos4  A;  J  v 


11. 


12. 


348                          Elementary  Calculus. 

/cot6  x  dx.  16.      I  sm    x  dx. 
J    COS4  X 

/*       •  A  /^sin  x  dx 

cos*  x  sm4  x  dx.  17.       I- 
J      COS3  X 

/"sin3  x  C   dx 

13.       I  — —  dx.  18.       I  — —  • 

J   COS^5  X  J   COS4  * 

/sin  *  /"          ^£ 

=•  dx.  19.       I  - 

cos  x  V  cos  *  J  cos4.*  sm2  * 

r  sin3  x  dx  r 

15.       I  ~7=         =-•  20.       I  sin4  *  cos4  * 

J  V  i  4-  cos  *  «y 

[.      /  sec3  *  d*         [set  sec  *  =  y]. 

/</*  r      dx 

sin  *  cos2  *  J    3  —  5  cos  * 

rco_s3_*^  2g>    r 

J    sin5  *  J 
^       C^cmxdx 
J  cot3  w* 

5.   r  ^ 

J  3  —  5  sin  * 

/; 


21. 


22. 


23- 
24. 

25- 

26.          . 

4-5  sin  2  * 

32.  I  emx  (sin  w*  —  cos 

33.  I  ex  cos  3  *  dx. 

34.  /  e3^  (cos  2  *  4-  sin  2  *)  dx. 


10  +  6  cos  5P 
sin  2  ^  dx. 


•  /elsi 

/  e2x  sin  4  * 

•7 


dx. 
cos  x  dx. 


dx. 


Elementary  Calculus.  349 

Integrate  the  following  by  multiple  angle  formulae: 
35.       /  sin2  x  cos4  x  dx.  37.       I  sin2  cos2  x  dx. 

36.  r./*  .  •      38.  f^dx. 

J  sm4  x  cos4  #  J  cos4  x 

MULTIPLE    INTEGRALS. 

ART.  101.  As  we  learned  that  a  given  function  may  have 
a  number  of  successive  derivatives,  it  immediately  follows 
that  a  multiple  derivative  admits  of  successive  integration, 
thus  recovering  the  lower  derivatives  and  eventually  the 
original  function.  This  process  is  indicated  by  repeating 
the  integral  sign,  thus, 


J  J  J 


Suppose  we  have,  for  example, 


This  is  what  is  known  as  a  differential  equation.  To  find 
the  relation  between  y  and  x  it  is  necessary  to  integrate 
three  times,  since  the  third  derivative  is  involved.  It 
follows  then,  that 

—  2  =  2  x2  dx  +  3  x  dxy 
dx2 

or    d  (^2\  =  2  x2  dx  +  3  x  dx. 
\dx2] 

Integrating, 

£2  =  2  fate  +  3  Cxdx=^  +  a*L+ci; 

dx2         J  J  32 


dy\  = 
\dxj 


350  Elementary  Calculus. 

Integrating, 

d2=2    Cy*dx  +  3    A 
dx~*J*  *J 


C2dx. 

6  2 

Integrating, 


Q,  C2,  and  C3  are  the  constants  of  integration  which  may 
be  determined  in  specific  cases  by  the  given  conditions  of 
the  problem.  This  process  is  useful  in  finding  the  equa- 
tions of  curves,  when  certain  attributes  expressed  in  terms  of 
their  derivatives  are  given,  for  example,  their  radii  of  curva- 
ture, although  a  general  application  to  this  end  requires 
a  general  knowledge  of  differential  equations. 

INTEGRATION    OF    A    TOTAL    DIFFERENTIAL. 

ART.  102.  Where  several  variables  are  involved  it  is 
necessary  to  reverse  the  process  of  partial  differentiation, 
thus  integrating  for  one  variable  at  a  time,  regarding  the 
others  as  constant.  In  the  case  of  a  function  of  two  vari- 
ables say,  z=  j  (x,  y),  the  expression  for  the  total  differ- 
ential is, 


Say  a  differential  is  given  in  the  form  P  dx  +  Q  dy, 
where  P  and  Q  are  functions  of  x  and  y.  If  the  function 
is  not  originally  in  this  form,  it  may  be  made  to  assume  it 
by  grouping. 


Elementary  Calculus.  351 

The  question  arises,  is  there  a  function  z,  of  x  and  y, 
which  will  have  the  expression  P  dx  +  Q  dy  for  its  differ- 
ential? 

x->.  *"\ 

Comparing  Pdx+Qdy  mth~dx+  ^-dy,  it  is  appar- 
ent that  if  there  is  such  a  function, 


Differentiating  these  equations  with  respect  to  y  and  x 
respectively, 


But 


.   3P  =  3Q 


And  when  this  is  true  the  function  z  exists,  not  otherwise. 

Example  :  3  x2  dx  +  3  y2  dy  —  3  a#  <£y  -  3  ^^  ^>  to 

/  (^,  y). 

Put  this  in  the  form  P  dx  +  Q  dy, 

(3  *2  -  3  <*?)  <&  +  (3  7s  -  3  «*)  <*?• 
Here  P  =  3  ^  ~  3  ^     Q  =  3  f  ~  3  ax- 


/B££  =  .S^and  z  exists. 
Since  P    =  3  x2  —  3  ay. 


352  Elementary  Calculus. 

Integrating  this  with  respect  to  x,  y  being  constant, 
zf  =  x3  —  3  axy  \_zp  means  partial  value  of  z]. 

Since  the  terms  in  Q,  which  contain  x,  have  already  been 
integrated  in  P,  as  will  be  evident  if  we  remember  how 
partial  differentiation  is  effected,  it  remains  only  to  inte- 
grate the  terms  in  Q  containing  y  alone,  with  respect  to  y. 

Since  Q  =  3  y2  —  2  ax,  the  integration  of  the  term  3  y2, 
containing  only  y,  gives  y. 

Adding  this  to  the  partial  integral  already  found  in  zp, 
the  total  integral  becomes, 

2=^  —  3  axy  +  y3. 

Hence  to  integrate  an  expression  of  the  form  P  dx  +  Q  dy, 
integrate  P  with  respect  to  x,  then  integrate  the  terms  in  Q 
not  containing  x,  and  add  the  results. 

DEFINITE    MULTIPLE    INTEGRALS. 

ART.  103.  Evidently  the  conception  of  multiple  integral 
may  include  definite  integration,  where  the  limits  of  inte- 
gration are  determined  for  each  variable  separately. 

/r     f*  \/r2  —  x2 
I  (x2  +  y2)dxdy 

0^/0 

means  that  the  definite  integral  of  this  expression  is  taken 
for  y  (x  remaining  constant)  between  the  limits  o  and 
vV  —  x2,  then  the  integral  of  this  result  with  respect  to  x, 
between  o  and  r. 

We  integrate  first  for  the  outside  differential. 

Thus, 

/V    /»VV2  —  a;2  /»r  /  *3\  vV'  —  x2 

I      I  (x2  +  y2)dxdy  =   I     !x2y+  2-)  dx 

JQ   JQ  Jo     \  3  /o 


Elementary  Calculus. 


353 


r4  .  ix~]  Tir* 
— sin-1-  =  — 
12  r  \ 


AREAS    AND    MOMENTS    OF    INERTIA. 

ART.  104.  The  determination  of  areas  comes  readily 
under  the  process  of  double  integration.  Take  the  circle 
(Fig.  30)  for  example.  Divide  the  circle  up  into  minute 


\ 


Fig.  30. 

squares,  by  lines  drawn  parallel  respectively  to  the  #-axis 
and  the  j-axis,  and  let  those  parallel  to  the  y-axis  be  at  a 
distance  A#  apart;  those  parallel  to  the  y-axis,  Ay  apart. 
Then  the  area  of  each  square  is  AJC  .  Ay.  The  sum  of  all 
these  squares  will  be  less  than  the  area  of  the  circle  by  the 
minute  spaces  bounded  by  the  sides  of  the  extreme  squares 
and  the  circumference.  But  as  Arc  and  Ay  approach  o, 
these  spaces  also  approach  o,  and  eventually  the  sum  of 
the  squares  represents  the  actual  area  of  the  circle,  that  is, 


354  Elementary  Calculus. 

when  A# .  A^  becomes  doc  •  dy.  We  have  learned  that 
definite  integration  is  a  summation,  hence  if  we  integrate 
along  a  line  parallel  to  the  ^-axis,  that  is  for  y,  we  get  a 
strip  parallel  to  the  #-axis,  and  then  integrating  parallel  to 
the  ;y-axis,  that  is  for  x,  we  sum  these  strips  and  hence  we 
get  the  circle  area.  Since  we  must  take  limits  for  y,  that 
will  apply  to  any  strip,  these  limits  or  rather  one  of  them 
will  be  variable,  and  should  be  a  function  of  x. 

Taking  the  origin  at  the  centre,  the  circle  equation  is 

y2  —   r2  —  X2, 


whence  y  =  \/r2  —  x2. 

Since  the  value  of  y  represents  any  point  on  the  circle,  it 
will  represent  the  distance  on  any  strip  from  the  #-axis, 
hence  starting  with  the  x-axis  and  integrating  upwards 
along  a  parallel  to  the  ;y-axis,  the  lower  limit  o  is  the  same 
for  all  strips  (the  starting  point  always  being  at  the  jg-axis) 
and  the  upper  limit  for  any  one  will  then  be  VV2  —  x2  (the 
outer  end  of  the  strip). 

Then  these  strips  are  integrated  parallel  to  the  #-axis, 
from  the  y-axis,  to  the  extreme  distance  of  the  last  one 
from  the  ;y-axis,  that  is,  r. 

We  express  all  this, 


r     /w/r2  —  a*  f*rT     ~|    W2  —  x* 

I  dxdy=         \y\ 

«/o  t/o   L    Jo 


=   C 
Jo 


—  ,  the  area  of  a  quadrant. 

4 

—  x  4  =  Tir2,  the  area  of  the  circle. 
4 


Elementary  Calculus.  355 

MOMENTS    OF    INERTIA. 

ART.  105.  The  moment  of  inertia  of  a  plane  area  about 
a  given  point  in  its  plane  is  defined  in  mechanics,  as  the 
sum  of  the  products  of  the  area  of  each  infinitesimal  portion 
by  the  square  of  its  distance  from  the  point. 

Taking  the  point  as  origin  and  laying  out  the  strips 
parallel  to  the  axes,  taking  the  axes  in  a  position  most 
convenient  for  laying  out  the  strips,  we  have  by  Analytic 
Geometry,  that  the  distance  of  any  point  (x,  y)  from  the 
point  (origin)  is 


Also  by  the  last  article  the  area  of  any  infinitesimal  square 
is  dx  dy. 

Since  an  infinitesimal  square  is  practically  a  point,  we 
have  then  the  moment  of  inertia  of  any  square  is 

(x2  +  y2)  dx  dy. 

Integrating  this  parallel  to  the  #-axis  with  proper  limits, 
determined  as  in  the  last  article,  and  then  parallel  to  the 
;y-axis  with  limits  indicating  the  extreme  of  area,  we  have 
the  required  sum.  Calling  the  moment  of  inertia,  I;  the 
limits  for  ^-integration,  (o,  a)  [where  a  is  a  function  of  x]; 
those  for  ^-integration,  (o,  Z>),  the  result  is  expressed, 


na 
(x 


This  was  illustrated  in  Art.  100.  The  same  process  may 
be  used  in  polar  co-ordinates  by  taking  radial  strips,  in- 
stead of  rectangular  ones. 


356  Elementary  Calculus. 

EXERCISE   XXI. 
By  double  integration  find  the  following: 

1.  The  area  between  y3  =  x  and  x3  =  y. 

2.  The  area  between  y2  =  8  x  and  x2  —  8  y. 

3.  The  area  between  y2  =  6  x  and  y2  =  10  x  —  x2. 

4.  Find  the  segment  of  the  circle  x2  +  y2  =  16  cut  off 
by  the  line  y  —  x  =  4. 

5.  Find  the  area  between  y2  =  2  px  and  the  line  y  =  2  x. 

6.  Find  the  moment  of  inertia  about  the  origin  of  the 
circle  (x  —  i)2  +  (y  —  z)2  =  9. 

7.  Find  the  moment  of  inertia  of  a  right  triangle,  about 
the  origin,  legs  of  length  6  in.  and  8  in.  respectively  forming 
the  axes. 

8.  Find  the  moment  of  inertia  of  the  area  in  Ex.  5. 

9.  Find  the  moment  of  inertia  of  the  segment  in  Ex.  4. 


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Imaginary   Substitution" 


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